tuyen tap de dh 2002-2012 theo chu de booklet
DESCRIPTION
Tuyen tap de dh 2002-2012 theo chu deTRANSCRIPT
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Tuyn tp cc thi i hc
2002-2012
theo ch
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Chng 7.Tch phn v ng dng 60
p s
7.1pi2
32+
1
4
7.2 12
(2 ln 3 3 ln 2)
7.3 I=
2
3+23
ln 2 + ln 3
7.4 343
+ 10 ln 35
7.5
3 + 2pi3
+ 12
ln 23
2+3
7.6 pi4
+ ln(2pi8
+22
)
7.7 (1)
7.8 (3 ln 3 2)
7.9 (e+ pi4 1)
7.10 I = (53e24
)
7.11 (5e4132
)
7.12 (32 ln 216
)
7.13 (ln(e2 + e+ 1) 2)
7.14 ( e22 1)
7.15 (12
ln 2)
7.16 (116135
)
7.17 (2 ln 2 1)
7.18 (ln 32)
7.19 (432
4)
7.20 (14(3 + ln 27
16))
7.21 (13
+ ln 32)
7.22 (14ln5
3)
7.23 (113 4 ln 2)
7.24 (3427
)
7.25 (23)
7.26 (12ln(2+
3) 10
93)
7.27 ( 815 pi
4)
7.28 (13
+ 12ln1+2e
3)
7.29 (2pi + 43)
7.30 (1096
)
7.31 ( e2 1)
7.32 (pi(5e32)27
)
Mc lc
1 Phng trnh-Bt PT-H PT-H BPT 31.1 Phng trnh v bt phng trnh . . . . . . . . . . . . . . . . . . 3
1.1.1 Phng trnh, bt phng trnh hu t v v t . . . . . . . 31.1.2 Phng trnh lng gic . . . . . . . . . . . . . . . . . . 41.1.3 Phng trnh,bt phng trnh m v logarit . . . . . . . . 8
1.2 H Phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Phng php hm s, bi ton cha tham s . . . . . . . . . . . . 12p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Bt ng thc 172.1 Bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Gi tr nh nht- Gi tr ln nht . . . . . . . . . . . . . . . . . . 182.3 Nhn dng tam gic . . . . . . . . . . . . . . . . . . . . . . . . . 20p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3 Hnh hc gii tch trong mt phng 223.1 ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 ng trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Cnic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 T hp v s phc 304.1 Bi ton m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
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4.2 Cng thc t hp . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 ng thc t hp khi khai trin . . . . . . . . . . . . . . . . . . . 314.4 H s trong khai trin nh thc . . . . . . . . . . . . . . . . . . . 324.5 S phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5 Kho st hm s 365.1 Tip tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Cc tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.3 Tng giao th . . . . . . . . . . . . . . . . . . . . . . . . . . 405.4 Bi ton khc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6 Hnh hc gii tch trong khng gian 446.1 ng thng v mt phng . . . . . . . . . . . . . . . . . . . . . 446.2 Mt cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Phng php ta trong khng gian . . . . . . . . . . . . . . . 51p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
7 Tch phn v ng dng 577.1 Tnh cc tch phn sau: . . . . . . . . . . . . . . . . . . . . . . . 577.2 Tnh din tch hnh phng c gii hn bi cc ng sau: . . . . 597.3 Tnh th tch khi trn xoay c to bi hnh phng (H) khi quay
quanh Ox. Bit (H) c gii hn bi cc ng sau: . . . . . . . 59p S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Chng 7.Tch phn v ng dng 59
Bi 7.22 (A-03).
I =
235
dx
xx2 + 4
.
Bi 7.23 (A-04).
I =
21
x
1 +x 1 dx.
Bi 7.24 (A-05).
I =
pi2
0
sin 2x+ sinx1 + 3 cosx
dx.
Bi 7.25 (A-06).
I =
pi2
0
sin 2xcos2 x+ 4 sin2 x
dx.
Bi 7.26 (A-08).
I =
pi6
0
tan4 x
cos 2xdx.
Bi 7.27 (A-09).
I =
pi2
0
(cos3 x 1) cos2 x dx.
Bi 7.28 (A-10).
I =
10
x2 + ex + 2x2ex
1 + 2exdx.
7.2 Tnh din tch hnh phng c gii hn bi ccng sau:
Bi 7.29 (B-02). y =
4 x24 v y = x2
4
2.
Bi 7.30 (A-02). y = |x2 4x+ 3| v y = x+ 3.Bi 7.31 (A-07). y = (e+ 1)x v y = (1 + ex)x.
7.3 Tnh th tch khi trn xoay c to bi hnhphng (H) khi quay quanh Ox. Bit (H) cgii hn bi cc ng sau:
Bi 7.32 (B-07). y = x lnx, y = 0 , x = e.
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Chng 7.Tch phn v ng dng 58
Bi 7.5 (B-11).
I =
pi3
0
1 + x sinx
cos2 xdx
Bi 7.6 (A-11).
I =
pi4
0
x sinx+ (x+ 1) cosx
x sinx+ cosxdx
Bi 7.7 (D-03).
I =
20
|x2 x| dx.
Bi 7.8 (D-04).
I =
32
ln(x2 x) dx.
Bi 7.9 (D-05).
I =
pi2
0
(esinx + cosx) cosx dx.
Bi 7.10 (D-06).
I =
10
(x 2)e2x dx.
Bi 7.11 (D-07).
I =
e1
x3 ln2 x dx.
Bi 7.12 (D-08).
I =
21
lnx
x3dx.
Bi 7.13 (D-09).
I =
31
dx
ex 1 .
Bi 7.14 (D-10).
I =
e1
(2x 3x
) lnx dx.
Bi 7.15 (B-03).
I =
pi4
0
1 2 sin2 x1 + sin 2x
dx.
Bi 7.16 (B-04).
I =
e1
1 + 3 lnx lnx
xdx.
Bi 7.17 (B-05).
I =
pi2
0
sin 2x cosx
1 + cos xdx.
Bi 7.18 (B-06).
I =
ln 5ln 3
dx
ex + 2ex 3 .
Bi 7.19 (B-08).
I =
pi4
0
sin(x pi4
)dx
sin 2x+ 2(1 + sin x+ cosx).
Bi 7.20 (B-09).
I =
31
3 + ln x
(x+ 1)2dx.
Bi 7.21 (B-10).
I =
e1
lnx
x(2 + lnx)2dx.
Chng 1
Phng trnh-Bt PT-H PT-HBPT
1.1 Phng trnh v bt phng trnh . . . . . . . . . . . . . . 31.1.1 Phng trnh, bt phng trnh hu t v v t . . . . . 31.1.2 Phng trnh lng gic . . . . . . . . . . . . . . . . 41.1.3 Phng trnh,bt phng trnh m v logarit . . . . . . 8
1.2 H Phng trnh . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Phng php hm s, bi ton cha tham s . . . . . . . . 12p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.1 Phng trnh v bt phng trnh
1.1.1 Phng trnh, bt phng trnh hu t v v t
Bi 1.1 (B-12). Gii bt phng trnh
x+ 1 +x2 4x+ 1 3x.
Bi 1.2 (B-11). Gii phng trnh sau:
3
2 + x 62 x+ 4
4 x2 = 10 3x (x R)
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Chng 1.Phng trnh-Bt PT-H PT-H BPT 4
Bi 1.3 (D-02). Gii bt phng trnh sau:
(x2 3x)
2x2 3x 2 0.Bi 1.4 (D-05). Gii phng trnh sau:
2
x+ 2 + 2
x+ 1x+ 1 = 4.
Bi 1.5 (D-06). Gii phng trnh sau:
2x 1 + x2 3x+ 1 = 0. (x R)Bi 1.6 (B-10). Gii phng trnh sau:
3x+ 16 x+ 3x2 14x 8 = 0.
Bi 1.7 (A-04). Gii bt phng trnh sau:2(x2 16)x 3 +
x 3 > 7 x
x 3 .
Bi 1.8 (A-05). Gii bt phng trnh sau:
5x 1x 1 > 2x 4.Bi 1.9 (A-09). Gii phng trnh sau:
2 3
3x 2 + 36 5x 8 = 0.Bi 1.10 (A-10). Gii bt phng trnh sau:
xx12(x2 x+ 1) 1.
1.1.2 Phng trnh lng gic
Bi 1.11 (D-12). Gii phng trnh sin 3x+ cos 3x sinx+ cosx =
2 cos 2x
Bi 1.12 (B-12). Gii phng trnh
2(cosx+
3 sinx) cosx = cosx
3 sinx+ 1.
Chng 7
Tch phn v ng dng
7.1 Tnh cc tch phn sau: . . . . . . . . . . . . . . . . . . . . 577.2 Tnh din tch hnh phng c gii hn bi cc ng sau: 597.3 Tnh th tch khi trn xoay c to bi hnh phng (H)
khi quay quanh Ox. Bit (H) c gii hn bi cc ngsau: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
p S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
7.1 Tnh cc tch phn sau:Bi 7.1 (D-12).
I =
pi4
0
x(1 + sin 2x)dx
Bi 7.2 (B-12).
I =
10
x3
x4 + 3x2 + 2dx.
Bi 7.3 (A-12).
I =
31
1 + ln(x+ 1)
x2dx
Bi 7.4 (D-11).
I =
40
4x 12x+ 1 + 2
dx
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Chng 6.Hnh hc gii tch trong khng gian 56
6.39 H(3; 0; 2); r = 4
6.40 x2 + y2 + (z + 2)2 = 25
6.41 V =a3
24
2
d =a6
6.42 V(SABH) = 7a311
96
6.43 V (S,ABC) = 13a27
4a = a
37
12
d [BC, SA] = 32HI = 3
2a42
12= a
428
6.44 V = 2a3
3;h = 6a7
6.45 V = 3a32
;h = a3
2
6.46 V(SMNCB) =
3a3;h = a1213
6.47 a3
2; a2
2
6.48 33a3
50
6.49 a3
6.50 a7
7
6.51 4a39
; 2a5
5
6.52 a314
48
6.53 a6; 90o
6.54 a
2
6.5526a3 tan
6.56 a32
36
6.57 a2
4
6.58 a33
3;55
6.59 9a3208
6.60 3a33
8; 7a12
6.61 a210
16
6.62 120o
6.633a3
12
6.643a3
96
6.65 a32
; 14
6.66 315a3
5
6.67 53a3
24; 23a19
Chng 1.Phng trnh-Bt PT-H PT-H BPT 5
Bi 1.13 (A-12). Gii phng trnh sau:
3 sin 2x+ cos 2x = 2 cos x 1Bi 1.14 (D-11). Gii phng trnh sau:
sin 2x+ 2 cosx sinx 1tanx+
3
= 0.
Bi 1.15 (B-11). Gii phng trnh sau:
sin 2x cosx+ sinx cosx = cos 2x+ sinx+ cosx
Bi 1.16 (A-11). Gii phng trnh
1 + sin 2x+ cos 2x
1 + cot2 x=
2 sinx sin 2x.
Bi 1.17 (D-02). Tm x thuc on [0; 14] nghim ng ca phng trnh:
cos 3x 4 cos 2x+ 3 cosx 4 = 0.Bi 1.18 (D-03). Gii phng trnh sau:
sin2 (x
2 pi
4) tan2 x cos2 x
2= 0.
Bi 1.19 (D-04). Gii phng trnh sau:
(2 cosx 1)(2 sinx+ cosx) = sin 2x sinx.Bi 1.20 (D-05). Gii phng trnh sau:
cos4 x+ sin4 x+ cos (x pi4
) sin (3x pi4
) 32
= 0.
Bi 1.21 (D-06). Gii phng trnh sau:
cos 3x+ cos 2x cosx 1 = 0.Bi 1.22 (D-07). Gii phng trnh sau:
(sinx
2+ cos
x
2)2
+
3 cosx = 2.
-
Chng 1.Phng trnh-Bt PT-H PT-H BPT 6
Bi 1.23 (D-08). Gii phng trnh sau:
2 sinx(1 + cos 2x) + sin 2x = 1 + 2 cosx.
Bi 1.24 (D-09). Gii phng trnh sau:
3 cos 5x 2 sin 3x cos 2x sinx = 0.Bi 1.25 (D-10). Gii phng trnh sau:
sin 2x cos 2x+ 3 sinx cosx 1 = 0.Bi 1.26 (B-02). Gii phng trnh sau:
sin2 3x cos2 4x = sin2 5x cos2 6x.Bi 1.27 (B-03). Gii phng trnh sau:
cotx tanx+ 4 sin 2x = 2sin 2x
.
Bi 1.28 (B-04). Gii phng trnh sau:
5 sinx 2 = 3(1 sinx) tan2 x.Bi 1.29 (B-05). Gii phng trnh sau:
1 + sin x+ cosx+ sin 2x+ cos 2x = 0.
Bi 1.30 (B-06). Gii phng trnh sau:
cotx+ sinx(1 + tan x tanx
2) = 4.
Bi 1.31 (B-07). Gii phng trnh sau:
2 sin2 2x+ sin 7x 1 = sin x.Bi 1.32 (B-08). Gii phng trnh sau:
sin3 x
3 cos3 x = sinx cos2 x
3 sin2 x cosx.
Bi 1.33 (B-09). Gii phng trnh sau:
sinx+ cosx sin 2x+
3 cos 3x = 2(cos 4x+ sin3 x).
Chng 6.Hnh hc gii tch trong khng gian 55
6.6 M(0; 1; 3) hayM(67; 47; 127
)
6.7 m = 12
6.8 k = 1
6.9 aba2+b2
; a = b = 2
6.10 15x+11y17z10 = 0;SOAB =5
6.11 A(1;4; 1); x11
= y23 =z35
6.12 x2
= y21 =z21
;M(1; 0; 4)
6.13 D(52; 12;1); x+3
1= y12 =
z11
6.14 xz22 = 0;M(4; 1; 1),M(7; 4; 4)
6.15 5
6.16 x+43
= y+22
= z41
6.17 x+ 3y + 5z 13 = 0M(0; 1;1), N(0; 1; 1)
6.18 x+ 2y 4z + 6 = 0;M(2; 3;7)
6.19 4x+2y+7z15 = 0; 2x+3z5 =0x+326
= y11
= z12
6.20 b = c = 12;M(1; 0; 0),M(2; 0; 0)
6.21 2x z = 0;H(2; 3; 4)
6.22 a2b4
; ab
= 1
6.23 30o, 26
3;
2
6.24 I(3; 5; 7)orI(3;7; 1)x = t, y = 1, z = 4 + t
6.25 122
2x y+ z 1 = 0; x 2y z+ 1 = 0
6.26 x27
= y1
= z+14
6.27 H(3; 1; 4);x 4y + z 3 = 0
6.28 M(0; 1;3)orM(1835
; 5335
; 335
)
6.29 16
6.30 (S) : (x2)2 + (y1)2 + (z3)2 =25.
6.31 (x+ 1)2 + (y + 1)2 + (z 2)2 =17
6.32 x2 + y2 + (z 3)2 = 83
6.33 (x5)2+(y11)2+(z2)2 = 1(x+ 1)2 + (y + 1)2 + (z + 1)2 = 1
6.34 x y + z = 0;x y z = 0
6.35 (x 1)2 + y2 + (z 1)2 = 1
6.36 x2 + y2 + z2 3x 3y 3z = 0H(2; 2; 2)
6.37 x2 + (y + 2)2 + z2 = 57625
;172
6.38 y 2z = 0;M(1;1;3)
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Chng 6.Hnh hc gii tch trong khng gian 54
Bi 6.61 (A-02). Cho hnh chp tam gic u S.ABC nh S, c di cnh ybng a. Gi M, N ln lt l trung im ca cc cnh SB v SC. Tnh theo a dintch tam gic AMN, bit rng mt phng (AMN) vung gc vi mt phng (SBC).
Bi 6.62 (A-03). Cho hnh lp phng ABCD.ABCD. Tnh s o gc phngnh din [B,AC,D].
Bi 6.63 (A-06). Cho hnh tr c cc y l hai hnh trn tm O v O, bn knhy bng chiu cao v bng a. Trn ng trn tm O ly im A, trn ng trntm O ly im B sao cho AB=2a. Tnh th tch khi t din OOAB.
Bi 6.64 (A-07). Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bnSAD l tam gic u v nm trong mt phng vung gc vi y. Gi M, N, P lnlt l trung im ca cc cnh SB, BC, CD. Chng minh rng AM vung gcvi BP v tnh th tch ca khi t din CMNP.
Bi 6.65 (A-08). Cho lng tr ABC.ABC c di cnh bn bng 2a, y ABCl tam gic vung ti A, AB=a, AC=a
3 v hnh chiu vung gc ca nh A
trn mt phng (ABC) l trung im ca cnh BC. Tnh theo a th tch khi chpAABC v tnh cosin ca gc gia hai ng thng AA v BC.
Bi 6.66 (A-09). Cho hnh chp S.ABCD c y ABCD l hnh thang vung tiA v D; AB=AD=2a, CD=a; gc gia hai mt phng (SBC) v (ABCD) bng 60o.Gi I l trung im ca cnh AD. Bit hai mt phng (SBI) v (SCI) cng vunggc vi mt phng (ABCD), tnh th tch khi chp S.ABCD theo a.
Bi 6.67 (A-10). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a.Gi M, N ln lt l trung im ca cc cnh AB v AD; H l giao im ca CNvi DM. Bit SH vung gc vi mt phng (ABCD) v SH=a
3. Tnh th tch
khi chp S.CDNM v tnh khong cch gia hai ng thng DM v SC theo a.
p s
6.1 M(1;1; 0) : M(73;5
3; 23).
6.2 6x+ 3y 4z + 12 = 0
6.3 x+12
= y+43
= z2
6.4 : x12
= y22
= z33
6.5 1. M = (4;7; 6); (5; 9;11)2. M(2; 1;5);M(14;35; 19)
Chng 1.Phng trnh-Bt PT-H PT-H BPT 7
Bi 1.34 (B-10). Gii phng trnh sau:
(sin 2x+ cos 2x) cosx+ 2 cos 2x sinx = 0.Bi 1.35 (A-02). Tm ngim thuc khong (0; 2pi) ca phng trnh:
5
(sinx+
cos 3x+ sin 3x
1 + 2 sin 2x
)= cos 2x+ 3.
Bi 1.36 (A-03). Gii phng trnh sau:
cotx 1 = cos 2x1 + tan x
+ sin2 x 12
sin 2x.
Bi 1.37 (A-05). Gii phng trnh sau:
cos2 3x cos 2x cos2 x = 0.Bi 1.38 (A-06). Gii phng trnh sau:
2(cos6 x+ sin6 x) sinx cosx2 2 sinx = 0.
Bi 1.39 (A-07). Gii phng trnh sau:
(1 + sin2 x) cosx+ (1 + cos2 x) sinx = 1 + sin 2x.
Bi 1.40 (A-08). Gii phng trnh sau:
1
sinx+
1
sin (x 3pi2
)= 4 sin (
7pi
4 x).
Bi 1.41 (A-09). Gii phng trnh sau:
(1 2 sinx) cosx(1 + 2 sinx)(1 sinx) =
3.
Bi 1.42 (A-10). Gii phng trnh sau:
(1 + sin x+ cos 2x) sin (x+pi
4)
1 + tan x=
12
cosx.
-
Chng 1.Phng trnh-Bt PT-H PT-H BPT 8
1.1.3 Phng trnh,bt phng trnh m v logarit
Bi 1.43 (D-11). Gii phng trnh sau:
log2(8 x2) + log 12(
1 + x+
1 x) 2 = 0 (x R)
Bi 1.44 (D-03). Gii phng trnh sau:
2x2x 22+xx2 = 3.
Bi 1.45 (D-06). Gii phng trnh sau:
2x2+x 4.2x2x 22x + 4 = 0.
Bi 1.46 (D-07). Gii phng trnh sau:
log2 (4x + 15.2x + 27) + 2 log2 (
1
4.2x 3) = 0.
Bi 1.47 (D-08). Gii bt phng trnh sau:
log 12
x2 3x+ 2x
0.
Bi 1.48 (D-10). Gii phng trnh sau:
42x+x+2 + 2x
3
= 42+x+2 + 2x
3+4x4 (x R)
Bi 1.49 (B-02). Gii bt phng trnh sau:
logx (log3 (9x 72)) 1.
Bi 1.50 (B-05). Chng minh rng vi mi x R, ta c:
(12
5)x
+ (15
4)x
+ (20
3)x
3x + 4x + 5x.
Khi no ng thc sy ra?
Bi 1.51 (B-06). Gii bt phng trnh sau:
log5 (4x + 144) 4 log2 5 < 1 + log5 (2x2 + 1).
Chng 6.Hnh hc gii tch trong khng gian 53
Bi 6.53 (B-02). Cho hnh lp phng ABCD.A1B1C1D1 c cnh bng a.a) Tnh theo a khong cch gia hai ng thng A1B v B1D.b) Gi M, N, P ln lt l trung im cc cnh BB1, CD, A1D1. Tnh gc gia haing thng MP v C1N.
Bi 6.54 (B-03). Cho hnh lng tr ng ABCD.ABCD c y ABCD l mthnh thoi cnh a, gc BAD = 60o. Gi M l trung im cnh AA v N l trungim cnh CC. Chng minh rng bn im B, M, D, N cng thuc mt phng.Hy tnh di cnh AA theo a t gic BMDN l hnh vung.
Bi 6.55 (B-04). Cho hnh chp t gic u S.ABCD c cnh y bng a, gcgia cnh bn v mt y bng (0o < < 90o). Tnh tan ca gc gia hai mtphng (SAB) v (ABCD) theo . Tnh th tch khi chp S.ABCD theo a v .
Bi 6.56 (B-06). Cho hnh chp S.ABCD c y ABCD l hnh ch nht viAB=a, AD=a
2, SA=a v SA vung gc vi mt phng (ABCD). Gi M, N ln
lt l trung im ca AD v SC, I l giao im ca BM v AC. Chng minh rngmt phng (SAC) vung gc vi mt phng (SMB). Tnh th tch khi t dinANIB.
Bi 6.57 (B-07). Cho hnh chp t gic u S.ABCD c y l hnh vung cnha. Gi E l im i xng ca D qua trung im ca SA, M l trung im caAE, N l trung im ca BC. Chng minh MN vung gc vi BD v tnh (theo a)khong cch gia hai ng thng MN v AC.
Bi 6.58 (B-08). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh 2a,SA=a, SB=a
3 v mt phng (SAB) vung gc vi mt phng y. Gi M, N ln
lt l trung im cc cnh AB, BC. Tnh theo a th tch ca khi chp S.BMDNv tnh cosin gc gia hai ng thng SM, DN.
Bi 6.59 (B-09). Cho hnh lng tr tam gic ABC.ABC c BB=a, gc giang thng BB v mt phng (ABC) bng 60o, tam gic ABC vung ti C vBAC = 60o. Hnh chiu vung gc ca im B ln mt phng (ABC) trng vitrng tm ca tam gic ABC. Tnh th tch khi t din AABC theo a.
Bi 6.60 (B-10). Cho hnh lng tr tam gic u ABC.ABC c AB=a, gc giahai mt phng (ABC) v (ABC) bng 60o. Gi G l trng tm tam gic ABC.Tnh th tch khi lng tr cho v tnh bn knh mt cu ngoi tip t dinGABC theo a.
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Chng 6.Hnh hc gii tch trong khng gian 52
Bi 6.45 (B-11). Cho lng trABCD.A1B1C1D1 c y ABCD l hnh ch nht.AB = a, AD = a
3. Hnh chiu vung gc ca im A1 trn mt phng (ABCD)
trng vi giao im AC v BD. Gc gia hai mt phng (ADD1A1) v (ABCD)bng 60o. Tnh th tch khi lng tr cho v khong cch t im B1 n mtphng (A1BD) theo a.
Bi 6.46 (A-11). Cho hnh chp S. ABC c y ABC l tam gic vung cn tiB, AB=BC=2a; hai mt phng (SAB) v (SAC) cng vung gc vi mt phng(ABC). Gi M l trung im ca AB; mt phng qua SM v song song vi BC,ct AC ti N. Bit gc gia hai mt phng (SBC) v (ABC) bng 60o. Tnh th tchkhi chp S. BCNM v khong cch gia hai ng thng AB v SN theo a.
Bi 6.47 (D-03). Cho hai mt phng (P) v (Q) vung gc vi nhau, c giao tuynl ng thng . Trn ly hai in A, B vi AB=a. Trong mt phng (P) lyin C, trong mt phng (Q) ly im D sao cho AC, BD cng vung gc vi v AC=BD=AB. Tnh bn knh mt cu ngoi tip t din ABCD v tnh khongcch t A n mt phng (BCD) theo a.
Bi 6.48 (D-06). Cho hnh chp tam gic S.ABC c y ABC l tam gic ucnh a, SA=2a v SA vung gc vi mt phng (ABC). Gi M, N ln lt l hnhchiu vung gc ca A trn cc ng thng SB v SC. Tnh th tch khi chpA.BCMN.
Bi 6.49 (D-07). Cho hnh chp S.ABCD c y l hnh thang, ABC = BAD =90o, BA=BC=a, AD=2a. Cnh bn SA vung gc vi dy v SA=a
2. Gi H l
hnh chiu vung gc ca A ln SB. Chng minh tam gic SCD vung v tnh(theo a) khong cch t H n mt phng (SCD).
Bi 6.50 (D-08). Cho lng tr ng ABC.ABC c y ABC l tam gic vung,AB=BC=a, cnh bn AA=a
2. Gi M l trung im cnh BC. Tnh theo a th
tch khi lng tr ABC.ABC v khong cch gia hai ng thng AM, BC.
Bi 6.51 (D-09). Cho lng tr ng ABC.ABC c y ABC l tam gic vungti B, AB=a, AA=2a, AC=3a. Gi M l trung im ca on thng AC, I l giaoim ca AM v AC. Tnh theo a th tch khi t din IABC v khong cch tim A n mt phng (IBC).
Bi 6.52 (D-10). Cho hnh chp S.ABCD c dy ABCD l hnh vung cnh a,cnh bn SA=a, hnh chiu vung gc ca nh S trn mt phng (ABCD) l im
H thuc on AC, AH=AC4. Gi CM l ng cao ca tam gic SAC. Chng
minh M l trung im ca SA v tnh th tch khi t din SMBC theo a.
Chng 1.Phng trnh-Bt PT-H PT-H BPT 9
Bi 1.52 (B-07). Gii phng trnh sau:
(
2 1)x + (
2 + 1)x 2
2 = 0.
Bi 1.53 (B-08). Gii bt phng trnh sau:
log0,7 (log6 (x2 + x
x+ 4)) < 0.
Bi 1.54 (A-06). Gii phng trnh sau:
3.8x + 4.12x 18x 2.27x = 0.
Bi 1.55 (A-07). Gii bt phng trnh sau:
2 log3 (4x 3) + log 13
(2x+ 3) 2.
Bi 1.56 (A-08). Gii phng trnh sau:
log2x1 (2x2 + x 1) + logx+1 (2x 1)2 = 4.
1.2 H Phng trnh
Bi 1.57 (D-12). Gii h phng trnh{xy + x 2 = 02x3 x2y + x2 + y2 2xy y = 0 ; (x; y R)
Bi 1.58 (A-12). Gii h phng trnh{x3 3x2 9x+ 22 = y3 + 3y2 9yx2 + y2 x+ y = 1
2
(x, y R).
Bi 1.59 (A-11). Gii h phng trnh:{5x2y 4xy2 + 3y3 2(x+ y) = 0xy(x2 + y2) + 2 = (x+ y)2
(x, y R)
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Chng 1.Phng trnh-Bt PT-H PT-H BPT 10
Bi 1.60 (D-02). Gii h phng trnh sau:23x = 5y2 4y
4x + 2x+1
2x + 2= y.
Bi 1.61 (D-08). Gii h phng trnh sau:{xy + x+ y = x2 2y2x
2y yx 1 = 2x 2y (x, y R).
Bi 1.62 (D-09). Gii h phng trnh sau:{x(x+ y + 1) 3 = 0(x+ y)2 5
x2+ 1 = 0
(x, y R).
Bi 1.63 (D-10). Gii h phng trnh sau:{x2 4x+ y + 2 = 02 log2 (x 2) log2 y = 0 (x, y R).
Bi 1.64 (B-02). Gii h phng trnh sau:{3x y = x y
x+ y =x+ y + 2.
Bi 1.65 (B-03). Gii h phng trnh sau:3y =
y2 + 2
x2
3x =x2 + 2
y2.
Bi 1.66 (B-05). Gii h phng trnh sau:{ x 1 +2 y = 1
3 log9 (9x2) log3 y3 = 3.
Bi 1.67 (B-08). Gii h phng trnh sau:{x4 + 2x3y + x2y2 = 2x+ 9x2 + 2xy = 6x+ 6
(x, y R).
Chng 6.Hnh hc gii tch trong khng gian 51
Bi 6.38 (B-07). Trong khng gian vi h ta Oxyz, cho mt cu(S): x2 +y2 +z22x+4y+2z3 = 0 v mt phng (P) : 2xy+2z14 = 0.1. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng trnc bn knh bng 3.2. Tm ta im M thuc mt cu (S) sao cho khong cch t M n mt phng(P) ln nht.
Bi 6.39 (A-09). Trong khng gian vi h ta Oxyz, cho mt phng(P):2x 2y z 4 = 0 v mt cu (S):x2 + y2 + z2 2x 4y 6z 11 = 0.Chng minh rng mt phng (P) ct mt cu (S) theo mt ng trn. Xc ta nh tm v bn knh ca ng trn .
Bi 6.40 (A-10). Trong khng gian ta Oxyz, cho im A(0;0;2) v ngthng
:x+ 2
2=y 2
3=z + 3
2. Tnh khong cch t A n . Vit phng trnh
mt cu tm A, ct ti hai im B v C sao cho BC=8.
6.3 Phng php ta trong khng gian
Bi 6.41 (D-12). Cho hnh hp ng ABCD.ABCD c y l hnh vung, tamgic AAC vung cn, AC = a. Tnh th tch khi t din ABBC v khong ccht im A n mt phng (BCD) theo a.
Bi 6.42 (B-12). Cho hnh chp tam gic u S.ABC vi SA = 2a, AB = a. Gi Hl hnh chiu vung gc ca A trn cnh SC. Chng minh SC vung gc vi mtphng (ABH). Tnh th tch ca khi chp S.ABH theo a.
Bi 6.43 (A-12). Cho hnh chp S.ABC c y l tam gic u cnh a. Hnh chiuvung gc ca S trn mt phng (ABC) l im H thuc cnh AB sao cho HA=2HB. Gc gia ng thng SC v mt phng (ABC) bng 60o. Tnh th tchca khi chp S.ABC v tnh khong cch gia hai ng thng SA v BC theo a.
Bi 6.44 (D-11). Cho hnh chp S.ABC c y ABC l tam gic vung ti B,BA = 3a, BC = 4a; mt phng (SBC) vung gc vi mt phng (ABC). Bit SB=2a
3v SBC = 30o. Tnh th tch khi chp S.ABC v khong cch t im Bn mt phng (SAC) theo a.
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Chng 6.Hnh hc gii tch trong khng gian 50
6.2 Mt cu
Bi 6.30 (D-12). Trong khng gian vi h ta Oxyz, cho mt phng (P): 2x+y2z + 10 = 0 v im I (2; 1; 3). Vit phng trnh mt cu tm I ct (P) theomt ng trn c bn knh bng 4.
Bi 6.31 (B-12). Trong khng gian vi h ta Oxyz, cho ng thng d :x 1
2=
y
1=
z
2 v hai im A(2;1;0), B(-2;3;2). Vit phng trnh mt cu iqua A,B v c tm thuc ng thng d.
Bi 6.32 (A-12). Trong khng gian vi h ta Oxyz, cho ng thng d:x+ 1
1=y
2=z 2
1v im I (0; 0; 3). Vit phng trnh mt cu (S) c tm I
v ct d ti hai im A, B sao cho tam gic IAB vung ti I.
Bi 6.33 (D-11). Trong khng gian vi h ta Oxyz, cho ng thng :x 1
2=y 3
4=z
1v mt phng (P): 2x y + 2z = 0. Vit phng trnh mt
cu c tm thuc ng thng , bn knh bng 1 v tip xc vi mt phng (P).
Bi 6.34 (A-11). Trong khng gian vi h ta Oxyz, cho mt cu(S) : x2 + y2 + z2 4x 4y 4z = 0 v im A(4; 4; 0). Vit phng trnh mtphng (OAB), bit im B thuc (S) v tam gic OAB u.
Bi 6.35 (D-04). Trong khng gian vi h ta Oxyz cho ba im A(2;0;1),B(1;0;0), C(1;1;1) v mt phng (P) : x + y + z 2 = 0. Vit phng trnh mtcu i qua ba im A, B, C, v c tm thuc mt phng (P).
Bi 6.36 (D-08). Trong khng gian vi h ta Oxyz, cho bn im A(3;3;0),B(3;0;3), C(0;3;3), D(3;3;3).1. Vit phng trnh mt cu i qua bn im A, B, C, D.2. Tm ta tm ng trn ngoi tip tam gic ABC.
Bi 6.37 (B-05). Trong khng gian vi h ta Oxyz, cho hnh lng tr ngABC.A1B1C1 vi A(0;3;0), B(4;0;0), C(0;3;0), B1(4;0;4).1. Tm ta cc nh A1, C1. Vit phng trnh mt cu c tm l A v tip xcvi mt phng (BCC1B1).2. Gi M l trung im ca A1B1. Vit phng trnh mt phng (P) i qua haiim A, M v song song vi BC1. Mt phng (P) ct ng thng A1C1 ti imN. Tnh di on MN.
Chng 1.Phng trnh-Bt PT-H PT-H BPT 11
Bi 1.68 (B-09). Gii h phng trnh sau:{xy + x+ 1 = 7yx2y2 + xy + 1 = 13y2
(x, y R).
Bi 1.69 (B-10). Gii h phng trnh sau:{log2 (3y 1) = x4x + 2x = 3y2.
Bi 1.70 (A-03). Gii h phng trnh sau: x1
x= y 1
y2y = x3 + 1.
Bi 1.71 (A-04). Gii h phng trnh sau: log 14 (y x) log41
y= 1
x2 + y2 = 25.
Bi 1.72 (A-06). Gii h phng trnh sau:{x+ y xy = 3x+ 1 +
y + 1 = 4.
Bi 1.73 (A-08). Gii h phng trnh sau:x2 + y + x3y + xy2 + xy = 5
4
x4 + y2 + xy(1 + 2x) = 54.
Bi 1.74 (A-09). Gii h phng trnh sau:{log2 (x
2 + y2) = 1 + log2 (xy)
3x2xy+y2 = 81.
Bi 1.75 (A-10). Gii h phng trnh sau:{(4x2 + 1)x+ (y 3)5 2y = 04x2 + y2 + 2
3 4x = 7.
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Chng 1.Phng trnh-Bt PT-H PT-H BPT 12
1.3 Phng php hm s, bi ton cha tham s
Bi 1.76 (D-11). Tm m h phng trnh sau c nghim{2x3 (y + 2)x2 + xy = mx2 + x y = 1 2m (x, y R)
Bi 1.77 (D-04). Tmm h phng trnh sau c nghim:{ x+y = 1
xx+ y
y = 1 3m.
Bi 1.78 (D-04). Chng minh rng phng trnh sau c ng mt nghim:
x5 x2 2x 1 = 0.
Bi 1.79 (D-06). Chng minh rng vi mi a > 0, h phng trnh sau c nghimduy nht: {
ex ey = ln (1 + x) ln (1 + y)y x = a.
Bi 1.80 (D-07). Tm gi tr ca tham sm phng trnh sau c nghim thc:x+
1
x+ y +
1
y= 5
x3 +1
x3+ y3 +
1
y3= 15m 10.
Bi 1.81 (B-04). Xc nhm phng trnh sau c nghim
m(
1 + x2
1 x2)
= 2
1 x4 +
1 + x2
1 x2.
Bi 1.82 (B-06). Tmm phng trnh sau c hai nghim thc phn bit:x2 +mx+ 2 = 2x+ 1.
Bi 1.83 (B-07). Chng minh rng vi mi gi tr dng ca tham s m, phngtrnh sau c hai nghim thc phn bit:
x2 + 2x 8 =m(x 2).
Chng 6.Hnh hc gii tch trong khng gian 49
Bi 6.25 (A-06). Trong khng gian vi h ta Oxyz, cho hnh lp phngABCD.ABCD vi A(0;0;0), B(1;0;0), D(0;1;0), A(0;0;1). Gi M v N ln ltl trung im ca AB v CD.1. Tnh khong cch gia hai ng thng AC v MN.2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc
bit cos =16.
Bi 6.26 (A-07). Trong khng gian vi h ta Oxyz, cho hai ng thng
d1 :x
2=y 11 =
z + 2
1v d2 :
x = 1 + 2ty = 1 + tz = 3.
1. Chng minh rng d1 v d2 cho nhau.2. Vit phng trnh ng thng d vung gc vi mt phng (P): 7x+y4z = 0v ct hai ng thng d1, d2.
Bi 6.27 (A-08). Trong khng gian vi h ta Oxyz, cho im A(2;5;3) vng thng d:
x 12
=y
1=z 2
2.
1. Tm ta hnh chiu vung gc ca im A trn ng thng d.2. Vit phng trnh mt phng () cha d sao cho khong cch t A n () lnnht.
Bi 6.28 (A-09). Trong khng gian vi h ta Oxyz, chomt phng (P):x2y+2z1 = 0 v hai ng thng 1 : x+ 1
1=y
1=z + 9
6,
2 :x 1
2=y 3
1=z + 1
2 . Xc nh ta im M thuc ng thng 1sao cho khong cch t M n ng thng 2 v khong cch t M n mtphng (P) bng nhau.
Bi 6.29 (A-10). Trong khng gian ta Oxyz, cho ng thng :x 1
2=
y
1=z + 2
1 v mt phng (P):x 2y + z = 0. Gi C l giao im ca vi (P),M l im thuc . Tnh khong cch t M n (P), bit MC=
6.
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Chng 6.Hnh hc gii tch trong khng gian 48
Bi 6.20 (B-10).1. Trong khng gian vi h ta Oxyz, cho cc im A(1;0;0), B(0;b;0),C(0;0;c), trong b, c dng v mt phng (P):y z + 1 = 0. Xc nh b vc, bit mt phng (ABC) vung gc vi mt phng (P) v khong cch t im O
n mt phng (ABC) bng1
3.
2. Trong khng gian vi h ta Oxyz, cho ng thng :x
2=y 1
1=z
2.
Xc nh ta im M trn trc honh sao cho khong cch t M n bngOM.
Bi 6.21 (A-02). Trong khng gian vi h ta Oxyz cho hai ng thng:
1 :
{x 2y + z 4 = 0x+ 2y 2z + 4 = 0 v 2 :
x = 1 + ty = 2 + tz = 1 + 2t
.
1. Vit phng trnh mt phng (P) cha ng thng 1 v song song vi ngthng 2.2. Cho im M(2;1;4). Tm ta im H thuc 2 sao cho on MH c dinh nht.
Bi 6.22 (A-03). Trong khng gian vi h ta Oxyz, cho hnh hp ch nhtABCD.ABCD c A trng vi gc ca h trc ta , B(a;0;0), D(0;a;0),A(0;0;b) (a> 0,b> 0). Gi M l trung im cnh CC.1. Tnh th tch khi t din BDAM theo a v b.2. Xc nh t s
a
b hai mt phng (ABD) v (MBD) vung gc vi nhau.
Bi 6.23 (A-04). Trong khng gian vi h ta Oxyz cho hnh chp S.ABCDc y ABCD l hnh thoi, AC ct BD ti gc ta O. Bit A(2;0;0), B(0;1;0),S(0;0;2
2). Gi M l trung im ca cnh SC.
1. Tnh gc v khong cch gia hai ng thng SA, BM.2. Gi s mt phng (ABM) ct ng thng SD ti im N. Tnh th tch khichp S.ABMN.
Bi 6.24 (A-05). Trong khng gian vi h trc ta Oxyz chong thng d:
x 11 =
y + 3
2=z 3
1v mt phng (P):2x+ y 2z + 9 = 0.
1. Tm ta im I thuc d sao cho khong cch t I n mt phng (P) bng 2.2. Tm ta giao im A ca ng thng d v mt phng (P). Vit phng trnhtham s ca ng thng nm trong mt phng (P), bit i qua A v vunggc vi d.
Chng 1.Phng trnh-Bt PT-H PT-H BPT 13
Bi 1.84 (A-02). Cho phng trnh:
log23 x+
log23 x+ 1 2m 1 = 0 (m l tham s).
1. Gii phng trnh khim = 2.2. Tmm phng trnh c t nht mt nghim thuc on [1; 3
3].
Bi 1.85 (A-07). Tmm phng trnh sau c nghim thc:
3x 1 +mx+ 1 = 2 4
x2 1.
Bi 1.86 (A-08). Tm cc gi tr ca tham s m phng trnh sau c ng hainghim thc phn bit:
4
2x+
2x+ 2 4
6 x+ 26 x = m (m R).
p s
1.1[
0 x 14
x 4
1.2 x = 65
1.3
x 12x = 2x 3
1.4 x = 3
1.5 x = 1 x = 22
1.6 x = 5
1.7 x > 1034
1.8 2 x < 10
1.9 x = 2
1.10 x = 35
2
1.11[x = pi
12+ k2pi
x = 7pi12
+ k2pi
1.12[x = 2pi
3+ k2pi
x = k2pi
1.13
x = pi2 + kpix = k2pix = 2pi
3+ k2pi
1.14 x = pi3
+ k2pi
1.15 cosx = 1; cosx = 12
1.16[x = pi
2+ kpi
x = pi4
+ k2pi
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Chng 1.Phng trnh-Bt PT-H PT-H BPT 14
1.17 x = pi2; x = 3pi
2; x = 5pi
2; x = 7pi
2
1.18
[x = pi + k2pi
x = pi4
+ kpi(k Z)
1.19[x = pi
3+ k2pi
x = pi4
+ kpi(k Z)
1.20 x =pi
4+ kpi (k Z)
1.21
[x = kpi
x = 2pi3
+ k2pi(k Z)
1.22[x = pi
2+ k2pi
x = pi6
+ k2pi(k Z)
1.23[x = 2pi
3+ k2pi
x = pi4
+ kpi(k Z)
1.24[x = pi
18+ k pi
3
x = pi6
+ k pi2
(k Z)
1.25[x = pi
6+ k2pi
x = 5pi6
+ k2pi(k Z)
1.26[x = kpi
9
x = kpi2
(k Z)
1.27 x = pi3
+ kpi (k Z)
1.28[x = pi
6+ k2pi
x = 5pi6
+ k2pi(k Z)
1.29[x = pi
4+ kpi
x = 2pi3
+ k2pi(k Z)
1.30[x = pi
12+ kpi
x = 5pi12
+ kpi(k Z)
1.31 x = pi8
+ k pi4
x = pi18
+ k 2pi3
x = 5pi18
+ k 2pi3
1.32[x = pi
4+ k pi
2
x = pi3
+ kpi(k Z)
1.33[x = pi
6+ k2pi
x = pi42
+ k 2pi7
(k Z)
1.34 x = pi4
+ k pi2
(k Z)
1.35[x = pi
3
x = 5pi3
1.36 x = pi4
+ kpi (k Z)
1.37 x = k pi2
(k Z)
1.38 x = 5pi4
+ k2pi (k Z)
1.39 x = pi4
+ kpix = pi
2+ k2pi
x = k2pi
1.40 x = pi4
+ kpix = pi
8+ kpi
x = 5pi8
+ kpi
1.41 x = pi18
+ k 2pi3
(k Z)
1.42[x = pi
6+ k2pi
x = 7pi6
+ k2pi(k Z)
1.43 x = 0
Chng 6.Hnh hc gii tch trong khng gian 47
2. Trong khng gian vi h ta Oxyz, cho hai ng thng 1 :
x = 3 + ty = tz = t
v 2 :x 2
2=y 1
1=z
2. Xc nh ta im M thuc 1 sao cho khong
cch t M n 2 bng 1.
Bi 6.15 (B-03). Trong khng gian vi h ta Oxyz, cho hai im A(2;0;0),B(0;0;8) v im C sao cho
AC =(0;6;0). Tnh khong cch t trung im I ca
BC n ng thng OA.
Bi 6.16 (B-04). Trong khng gian vi h ta Oxyz, cho im A(4;2;4) v
ng thng d:
x = 3 + 2ty = 1 tz = 1 + 4t
. Vit phng trnh ng thng i qua im
A, ct v vung gc vi ng thng d.
Bi 6.17 (B-06). Trong khng gian vi h ta Oxyz, cho im A(0;1;2) v haing thng:
d1 :x
2=y 1
1=z + 1
1 , d2 :
x = 1 + ty = 1 2tz = 2 + t.
1. Vit phng trnh mt phng (P) qua A, ng thi song song vi d1 v d2.2. Tm ta cc im M thuc d1, N thuc d2 sao cho ba im A, M, N thnghng.
Bi 6.18 (B-08). Trong khng gian vi h ta Oxyz, cho ba im A(0;1;2),B(2;2:1), C(2;0;1).1. Vit phng trnh mt phng i qua ba im A, B, C.2. Tm ta imM thuc mt phng 2x+2y+z3 = 0 sao choMA=MB=MC.Bi 6.19 (B-09).1. Trong khng gian vi h ta Oxyz, cho t din ABCD c cc nh A(1;2;1),B(2;1;3), C(2;1;1) v D(0;3;1). Vit phng trnh mt phng (P) i qua A, Bsao cho khong cch t C n (P) bng khong cch t D n (P).2. Trong khng gian vi h ta Oxyz, cho mt phng (P): x 2y+ 2z 5 = 0v hai im A(3;0;1), B(1;1;3). Trong cc ng thng i qua A v song songvi (P), hy vit phng trnh ng thng m khong cch t B n ng thng nh nht.
-
Chng 6.Hnh hc gii tch trong khng gian 46
Bi 6.10 (D-05). Trong khng gian vi h ta Oxyz cho hai ng thng
d1 :x 1
3=y + 2
1 =z + 1
2v d2 :
{x+ y z 2 = 0x+ 3y 12 = 0.
a) Chng minh rng d1 v d2 song song vi nhau. Vit phng trnh mt phng(P) cha c hai ng thng d1 v d2.b) Mt phng ta Oxy ct c hai ng thng d1, d2 ln lt ti cc im A, B.Tnh din tch tam gic OAB (O l gc ta ).
Bi 6.11 (D-06). Trong khng gian vi h ta Oxyz, cho im A(1;2;3) v haing thng:
d1 :x 2
2=y + 2
1 =z 3
1, d2 :
x 11 =
y 12
=z + 1
1.
1. Tm ta im A i xng vi im A qua ng thng d1.2. Vit phng trnh ng thng i qua A, vung gc vi d1 v ct d2.
Bi 6.12 (D-07). Trong khng gian vi h ta Oxyz, cho hai im A(1;4;2),B(1;2;4) v ng thng : x 11 =
y + 2
1=z
2.
1. Vit phng trnh ng thng d i qua trng tm G ca tam gic OAB vvung gc mt phng (OAB).2. Tm ta im M thuc ng thng sao cho MA2+MB2 nh nht.
Bi 6.13 (D-09).1. Trong khng gian vi h ta Oxyz,cho cc im A(2;1;0), B(1;2;2), C(1;1;0)v mt phng (P): x+ y+ z 20 = 0. Xc nh ta im D thuc ng thngAB sao cho ng thng CD song song vi mt phng (P).
2. Trong khng gian vi h ta Oxyz, cho ng thng :x+ 2
1=y 2
1=
z
1 v mt phng (P):x+ 2y 3z+ 4 = 0. Vit phng trnh ng thng d nmtrong (P) sao cho d ct v vung gc vi ng thng .
Bi 6.14 (D-10).1. Trong khng gian vi h ta Oxyz, cho hai mt phng (P): x+y+ z3 = 0v (Q):x y + z 1 = 0. Vit phng trnh mt phng (R) vung gc vi (P) v(Q) sao cho khong cch t O n (R) bng 2.
Chng 1.Phng trnh-Bt PT-H PT-H BPT 15
1.44[x = 1x = 2
1.45 x = 0 x = 1
1.46 x = log2 3
1.47 S = [22; 1) (2; 2 +2]
1.48 x = 1 x = 2
1.49 log9 73 < x 2
1.50 x = 0
1.51 2 < x < 4
1.52 x = 1 x = 1
1.53 S = (4;3) (8; +)
1.54 x = 1
1.55 34< x 3
1.56 x = 2 x = 54
1.57
(x; y) = (1; 1)(1+52
;
5)
(15
2;5)
1.58 (x; y) =(32;1
2
);(12; 3
2
)1.59 (1; 1); (1;1); (2
25
;25);
(225
;25)
1.60
{x = 0
y = 1{x = 2
y = 4
1.61 (x; y) = (5; 2)
1.62 (x; y) = (1; 1); (2;32
)
1.63 (x; y) = (3; 1)
1.64 (x; y) = (1; 1); (32; 12)
1.65 x = y = 1
1.66 (x; y) = (1; 1); (2; 2)
1.67 (x; y) = (4; 174
)
1.68 (x; y) = (1; 13); (3; 1)
1.69 (x; y) = (1; 12)
1.70 (x; y) = (1; 1); (1+5
2; 1+
5
2)
(15
2; 1
5
2)
1.71 (x; y) = (3; 4)
1.72 (x; y) = (3; 3)
1.73 (x; y) = ( 3
54; 3
2516
) = (1;32)
1.74 x = y = 2x = y = 2
1.75 (x; y) = (12; 2)
1.76 m 23
2
1.77 0 m 14
-
Chng 1.Phng trnh-Bt PT-H PT-H BPT 16
1.78 f(x) = vt b trn[1; +)
1.80
[ 74 m 2
m 22
1.81
2 1 m 1
1.82 m 92
1.83
1.84 1.x = 33
2.0 m 2
1.85 1 < m 13
1.86 2
6 + 2 4
6 m < 32 + 6
Chng 6.Hnh hc gii tch trong khng gian 45
Bi 6.4 (D-11). Trong khng gian vi h ta Oxyz, cho im A(1; 2; 3) v
ng thng d :x+ 1
2=y
1=z 32 . Vit phng trnh ng thng i qua
im A, vung gc vi ng thng d v ct trc Ox.
Bi 6.5 (B-11).
1. Trong khng gian h to Oxyz, cho ng thng
:x 2
1=y + 1
2 =z
1 v mt phng (P): x + y + z 3 = 0. Gi Il giao im ca v (P). Tm ta im M thuc (P) sao cho MI vunggc vi v MI =4
14.
2. Trong khng gian vi h to Oxyz, cho ng thng
:x+ 2
1=y 1
3=z + 5
2 v hai im A(2; 1; 1);B(3;1; 2). Tmta im M thuc ng thng sao cho tam gic MAB c din tchbng 3
5.
Bi 6.6 (A-11). Trong khng gian vi h ta Oxyz, cho hai imA(2; 0; 1), B(0;2; 3) v mt phng (P ) : 2x y z + 4 = 0. Tm ta imM thuc (P ) sao choMA = MB = 3.
Bi 6.7 (D-02). Trong khng gian vi h ta Oxyz, cho mt phng(P) : 2x y + 2 = 0 v ng thngdm :
{(2m+ 1)x+ (1m)y +m 1 = 0mx+ (2m+ 1)z + 4m+ 2 = 0
(m l tham s).Xc nhm ng thng dm song song vi mt phng (P).
Bi 6.8 (D-03). Trong khng gian vi h ta Oxyz, cho mt phng
(P) : x y2z+ 5 = 0 v ng thng dk :{x+ 3ky z + 2 = 0kx y + z + 1 = 0 (k l tham
s).Xc nh k ng thng dk vung gc vi mt phng (P).
Bi 6.9 (D-04). Trong khng gian vi h ta Oxyz cho hnh lng tr ngABC.A1B1C1. Bit A(a;0;0), B(a;0;0), C(0;1;0), B1(a;0;b), a> 0, b> 0.a) Tnh khong cch gia hai ng thng B1C v AC1 theo a, b.b) Cho a, b thay i, nhng lun tha mn a+b = 4. Tm a, b khong cch giahai ng thng B1C v AC1 ln nht.
-
Chng 6
Hnh hc gii tch trong khng gian
6.1 ng thng v mt phng . . . . . . . . . . . . . . . . . . 446.2 Mt cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Phng php ta trong khng gian . . . . . . . . . . . . 51p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.1 ng thng v mt phng
Bi 6.1 (D-12). Trong khng gian vi h ta Oxyz, cho ng thng d :x 1
2=
y + 1
1 =z
1v hai im A (1; -1; 2), B (2; -1; 0). Xc nh ta
im M thuc d sao cho tam gic AMB vung ti M.
Bi 6.2 (B-12). Trong khng gian vi h ta Oxyz, cho A(0;0;3), M(1;2;0).Vit phng trnh mt phng (P) qua A v ct cc trc Ox, Oy ln lt ti B, Csao cho tam gic ABC c trng tm thuc ng thng AM.
Bi 6.3 (A-12). Trong khng gian vi h ta Oxyz, cho ng thngd:x+ 1
2=y
1=z 2
1, mt phng (P) :x + y 2z + 5 = 0 v im A (1; -1;
2). Vit phng trnh ng thng ct d v (P) ln lt ti M v N sao cho A ltrung im ca on thng MN.
Chng 2
Bt ng thc
2.1 Bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Gi tr nh nht- Gi tr ln nht . . . . . . . . . . . . . . . 182.3 Nhn dng tam gic . . . . . . . . . . . . . . . . . . . . . . 20p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.1 Bt ng thc
Bi 2.1 (A-09). Chng minh rng vi mi s thc dng x, y, ztha mn x(x+ y + z) = 3yz, ta c:
(x+ y)3 + (x+ z)3 + 3(x+ y)(x+ z)(y + z) 5(y + z)3.Bi 2.2 (A-05). Cho x, y, z l cc s dng tha mn
1
x+
1
y+
1
z= 4. Chng
minh rng1
2x+ y + z+
1
x+ 2y + z+
1
x+ y + 2z 1.
Bi 2.3 (A-03). Cho x, y, z l ba s dng v x+ y + z 1. Chng minh rngx2 +
1
x2+
y2 +
1
y2+
z2 +
1
z2
82.
-
Chng 2.Bt ng thc 18
Bi 2.4 (D-07). Cho a b > 0. Chng minh rng :(
2a +1
2a
)b(
2b +1
2b
)a.
Bi 2.5 (D-05). Cho cc s dng x, y, z tha mn xyz = 1. Chng minh rng1 + x3 + y3
xy+
1 + y3 + z3
yz+
1 + z3 + x3
zx 3
3.
Khi no ng thc xy ra?
2.2 Gi tr nh nht- Gi tr ln nht
Bi 2.6 (D-12). Cho cc s thc x, y tha mn (x4)2 + (y4)2 + 2xy 32. Tmgi tr nh nht ca biu thc A = x3 + y3 + 3(xy1)(x+ y2).
Bi 2.7 (B-12). Cho cc s thc x, y, z tha mn cc iu kin x+ y + z = 0 v
x2 + y2 + z2 = 1.
Tm gi tr ln nht ca biu thc
P = x5 + y5 + z5.
Bi 2.8 (A-12). Cho cc s thc x, y, z tha mn iu kin x + y + z = 0. Tmgi tr nh nht ca biu thc
P = 3|xy| + 3|yz| + 3|zx|
6x2 + 6y2 + 6z2
Bi 2.9 (B-11). Cho a v b l cc s thc dng tha mn2(a2 + b2) + ab = (a + b)(ab + 2). Tm gi tr nh nht ca biu thc
P= 4(a3
b3+b3
a3
) 9
(a2
b2+b2
a2
).
Bi 2.10 (A-11). Cho x, y, z l ba s thc thuc on [1; 4] v x y, x z. Tmgi tr nh nht ca biu thc
P =x
2x+ 3y+
y
y + z+
z
z + x
.
Chng 5.Kho st hm s 43
5.24 I(13< m < 1,m 6= 0); II(m =
1)
5.25 I(0 < m < 1); II(m = 26)
5.26 m = 2
5.27 12< m < 0
5.28 m = 15
2
5.29 4 < m < 5
5.30 14< m < 1,m 6= 0
5.31 m = 0 orm = 2
5.32 m > 0
5.33 m = 1
-
Chng 5.Kho st hm s 42
Bi 5.32 (B-03). Cho hm s : y = x3 3x2 +m (1) (m l tham s).1. Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gcta .2. Kho st s bin thin v v th ca hm s (1) ng vi m= 2.
Bi 5.33 (A-08). Cho hm s y =mx2 + (3m2 2)x 2
x+ 3m(1), vi m l tham
s thc.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm cc gi tr ca tham s m gc gia hai ng tim cn ca th hms (1) bng 45o.
p s
5.1 m = 1
5.2 1 + 4 ln 43;m 6= 1
5.3 m = 4
5.4 M(12;2);M(1; 1)
5.5 y = 6x+ 10
5.6 y = x+ 83
5.7 y = x+225; y = x225
5.8 y = 24x+ 15; y = 154x 21
4
5.9 y = x 2
5.10 m = 23
5.11 m = 2
5.12 m = 0
5.13 m = 2 22
5.14 m < 3 or 0 < m < 3
5.15 M(2;m 3);N(0;m+ 1)
5.16 m = 12
5.17 1 < k < 3, k 6= 0, k 6= 2y = 2xm2 +m
5.18 m = 1
5.19 m = 4 26
5.20 k = 3
5.21 m > 1
5.22 m > 154,m 6= 24
5.23
Chng 2.Bt ng thc 19
Bi 2.11 (D-11). Tm gi tr nh nht v gi tr ln nht ca hm s y =2x2 + 3x+ 3
x+ 1trn on [0; 2].
Bi 2.12 (A-07). Cho x, y, z l cc s thc dng thay i v tha mn iu kinxyz = 1. Tm gi tr nh nht ca biu thc:
P =x2(y + z)
yy + 2z
z
+y2(z + x)
zz + 2x
x
+z2(x+ y)
xx+ 2y
y.
Bi 2.13 (A-06). Cho hai s thc x 6= 0, y 6= 0 thay i v tha mn iu kin:
(x+ y)xy = x2 + y2 xy.
Tm gi tr ln nht ca biu thc
A =1
x3+
1
y3.
Bi 2.14 (B-10). Cho cc s thc khng m a, b, c tha mn a + b + c = 1. Tmgi tr nh nht ca biu thc
M = 3(a2b2 + b2c2 + c2a2) + 3(ab+ bc+ ca) + 2a2 + b2 + c2.
Bi 2.15 (B-09). Cho cc s thc x, y thay i v tha mm (x+ y)3 + 4xy 2.Tm gi tr nh nht ca biu thc
A = 3(x4 + y4 + x2y2) 2(x2 + y2) + 1.
Bi 2.16 (B-08). Cho hai s thc x, y thay i v tha mn h thc x2 + y2 = 1.Tm gi tr ln nht v gi tr nh nht ca biu thc
P =2(x2 + 6xy)
1 + 2xy + 2y2.
Bi 2.17 (B-07). Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nhtca biu thc:
P = x
(x
2+
1
yz
)+ y
(y
2+
1
zx
)+ z
(z
2+
1
xy
).
-
Chng 2.Bt ng thc 20
Bi 2.18 (B-06). Cho x, y l cc s thc thay i. Tm gi tr nh nht ca biuthc:
A =
(x 1)2 + y2 +
(x+ 1)2 + y2 + |y 2|.Bi 2.19 (B-03). Tm gi tr ln nht v gi tr nh nht ca hm sy = x+
4 x2.
Bi 2.20 (D-10). Tm gi tr nh nht ca hm s
y =x2 + 4x+ 21
x2 + 3x+ 10.
Bi 2.21 (D-09). Cho cc s thc khng m x, y thay i v tha mn x+ y = 1.Tm gi tr ln nht v gi tr nh nht ca biu thc
S = (4x2 + 3y)(4y2 + 3x) + 25xy.
Bi 2.22 (D-08). Cho x, y l hai s thc khng m thay i. Tm gi tr ln nhtv gi tr nh nht ca biu thc
P =(x y)(1 xy)(1 + x)2(1 + y)2
.
Bi 2.23 (D-03). Tm gi tr ln nht v gi tr nh nht ca hm s y =x+ 1x2 + 1
trn on [1; 2].
2.3 Nhn dng tam gic
Bi 2.24 (A-04). Cho tam gic ABC khng t tha mn iu kin
cos 2A+ 2
2 cosB + 2
2 cosC = 3.
Tnh ba gc ca tam gic ABC.
p s
Chng 5.Kho st hm s 41
2. Vi gi tr no ca m, phng trnh x2|x2 2| = m c ng 6 nghim thcphn bit?II. Tm cc gi tr ca tham s m ng thng y = x + m ct th hm sy =
x2 1x
ti hai im phn bit A, B sao cho AB= 4.
Bi 5.26 (B-10). Cho hm s y =2x+ 1
x+ 1.
1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm m ng thng y = 2x+m ct th (C) ti hai im phn bit A, Bsao cho tam gic OAB c din tch bng
3 (O l gc ta ).
Bi 5.27 (A-03). Cho hm s y =mx2 + x+m
x 1 (1) (ml tham s).1. Kho st s bin thin v v th hm s (1) khim = 1.2. Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im c honh dng.
Bi 5.28 (A-04). Cho hm s y =x2 + 3x 3
2(x 1) (1).1. Kho st hm s (1).2. Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao choAB= 1.
Bi 5.29 (A-06).1. Kho st s bin thin v v th hm s y = 2x3 9x2 + 12x 4.2. Tmm phng trnh sau c 6 nghim phn bit: 2|x3| 9x2 + 12|x| = m.Bi 5.30 (A-10). Cho hm s y = x3 2x2 + (1 m)x + m (1), m l tham sthc.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm m th ca hm s (1) ct trc honh ti 3 im phn bit c honh x1, x2, x3 tha mn iu kin x21 + x
22 + x
23 < 4.
5.4 Bi ton khc
Bi 5.31 (D-04). Cho hm s : y = x3 3mx2 + 9x + 1 (1) (m l thams).1. Kho st hm s (1) ng vim = 2.2. Tm m im un ca th hm s (1) thuc ng thng y = x+ 1.
-
Chng 5.Kho st hm s 40
5.3 Tng giao th
Bi 5.20 (D-11). Cho hm s y =2x+ 1
x+ 1
1. Kho st s bin thin v v th (C) ca hm s cho
2. Tm k ng thng y = kx+ 2k+ 1 ct th (C) ti hai im phn bitA, B sao cho khong cch t A v B n trc honh bng nhau.
Bi 5.21 (D-03).
1. Kho st s bin thin v v th ca hm s y =x2 2x+ 4
x 2 (1).2. Tm m ng thng dm: y = mx + 2 2m ct th hm s (1) ti haiim phn bit.
Bi 5.22 (D-06). Cho hm s : y = x3 3x+ 2.1. Kho st s bin thin v v th (C) ca hm s cho.2. Gi d l ng thng i qua im A(3; 20) v c h s gc l m. Tm m ng thng d ct th (C) ti 3 im phn bit.
Bi 5.23 (D-08). Cho hm s : y = x3 3x2 + 4 (1).1. Kho st s bin thin v v th ca hm s (1).2. Chng minh rng mi ng thng i qua im I(1; 2) vi h s gc k (k> 3)u ct th ca hm s (1) ti ba im phn bit I, A, B ng thi I l trungim ca on thng AB.
Bi 5.24 (D-09).I. Cho hm s y = x4 (3m+ 2)x2 + 3m c th l (Cm), m l tham s.1. Kho st s bin thin v v th ca hm s cho khi m= 0.2. Tm m ng thng y = 1 ct th (Cm) ti 4 im phn bit u chonh nh hn 2.II. Tm cc gi tr ca tham sm ng thng y = 2x+m ct th hm sy =
x2 + x 1x
ti hai im phn bit A, B sao cho trung im ca on thngAB thuc trc tung.
Bi 5.25 (B-09).I. Cho hm s y = 2x4 4x2 (1).1. Kho st s bin thin v v th ca hm s (1).
Chng 2.Bt ng thc 21
2.6 Amin = 1755
4
2.7 P =5
6
36
2.8 Pmin = 3
2.9 minP = 234
2.10 Pmin = 3433
2.11 GTLN l 173;GTNN
l 3
2.12 Pmin = 2
2.13 Amax = 16
2.14 Mmin = 2
2.15 Amin =9
16
2.16 Pmax = 3;Pmin =6
2.17 Pmin =9
2
2.18 Amin = 2 +
3
2.19 max[2;2]
y = 2
2
min[2;2]
y = 2
2.20 ymin =
2
2.21 Smax = 252 ;Smin =19116
2.22 Pmin =1
4;Pmax =
14
2.23 ymax =
2; ymin =0
2.24 A = 90o;B = C =45o
-
Chng 3
Hnh hc gii tch trong mt phng
3.1 ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 ng trn . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Cnic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.1 ng thng
Bi 3.1 (D-12). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD.Cc ng thng AC v AD ln lt c phng trnh l x+3y = 0 v xy+4 = 0;ng thng BD i qua im M (1
3; 1). Tm ta cc nh ca hnh ch nht
ABCD.
Bi 3.2 (A-12). Trong mt phng vi h ta Oxy, cho hnh vung ABCD. GiM l trung im ca cnh BC, N l im trn cnh CD sao cho CN = 2ND. Gi sM(112
; 12
)v ng thng AN c phng trnh 2x y 3 = 0. Tm ta im
A.
Bi 3.3 (D-11). Trongmt phng ta Oxy, cho tam gic ABC c nhB(4; 1),trng tm G(1; 1) v ng thng cha phn gic trong ca gc A c phng trnhx y 1 = 0. Tm ta cc nh A v C.
Chng 5.Kho st hm s 39
Bi 5.14 (B-02). Cho hm s : y = mx4 + (m2 9)x2 + 10 (1) (m ltham s).1. Kho st s bin thin v v th ca hm s (1) ng vi m= 1.2. Tm m hm s (1) c ba im cc tr.
Bi 5.15 (B-05). Gi (Cm) l th ca hm s y =x2 + (m+ 1)x+m+ 1
x+ 1(*)
(m l tham s).1. Kho st v v th hm s (*) khi m= 1.2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, imcc tiu v khong cch gia hai im bng
20.
Bi 5.16 (B-07). Cho hm s: y = x3 + 3x2 + 3(m2 1)x 3m2 1 (1), m ltham s.1. Kho st s bin thin v v th hm s (1) khi m= 1.2. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm s(1) cch u gc ta O.
Bi 5.17 (A-02). Cho hm s: y = x3 + 3mx2 + 3(1m2)x+m3 m2 (1)(m l tham s).1. Kho st s bin thin v v th ca hn s (1) khim = 1.2. Tm k phng trnh: x3 + 3x2 + k3 3k2 = 0 c ba nghim phnbit.3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1).
Bi 5.18 (A-05). Gi(Cm) l th ca hm s y = mx+1
x(*) (m l tham
s).
1. Kho st s bin thin v v th ca hm s (*) khim =1
4.
2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm) n
tim cn xin ca (Cm) bng12.
Bi 5.19 (A-07). Cho hm s y =x2 + 2(m+ 1)x+m2 + 4m
x+ 2(1),m l tham
s.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cng vi gc ta O to thnh mt tam gic vung ti O.
-
Chng 5.Kho st hm s 38
Bi 5.9 (A-09). Cho hm s y =x+ 2
2x+ 3(1).
1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct trchonh, trc tung ln lt ti hai im phn bit A, B v tam gic OAB cn ti gcta O.
5.2 Cc tr
Bi 5.10 (D-12). Cho hm s y = 23x3mx22(3m21)x + 2
3(1), m l tham s
thc.
1. Kho st s bin thin v v th ca hm s (1) khi m = 1.
2. Tmm hm s (1) c hai im cc tr x1 vx2 sao cho x1.x2+2(x1+x2) =1
Bi 5.11 (B-12). Cho hm s y = x3 3mx2 + 3m2, (1),m l tham s thc.
1. Kho st s bin thin v v th ca hm s (1) khi m = 1.
2. Tm m th hm s (1) c hai im cc tr A v B sao cho tam gicOAB c din tch bng 48.
Bi 5.12 (A-12). Cho hm s y = x4 2(m + 1)x2 + m2, (1) ,vi m l thams.
1. Kho st s bin thin v v th hm s (1) khim = 0.
2. Tm m th hm s (1) c ba im cc tr to thnh ba nh ca mttam gic vung.
Bi 5.13 (B-11). Cho hm s y = x4 2(m+ 1)x2 +m, (1), m l tham s.
1. Kho st s bin thin v v th hm s (1) khi m = 1.
2. Tm m th hm s (1) c ba im cc tr A, B, C sao cho OA = BC, Ol gc ta , A l cc tr thuc trc tung, B v C l hai im cc tr cn li.
Chng 3.Hnh hc gii tch trong mt phng 23
Bi 3.4 (B-11). Trong mt phng to Oxy, cho hai ng thng : xy4 =0 v d : 2x y 2 = 0. Tm ta im N thuc ng thng d sao cho ngthng ON ct ng thng ti im M tha mn OM.ON = 8.
Bi 3.5 (A-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC cn ti A c nh A(6;6), ng thng i qua trung im ca cc cnhAB v AC c phng trnh x+ y 4 = 0. Tm ta cc nh B v C, bit imE(1;-3) nm trn ng cao i qua nh C ca tam gic cho.
Bi 3.6 (A-09). Trong mt phng vi h ta cac vung gc Oxy, cho hnhch nht ABCD c im I(6;2) l giao im ca hai ng cho AC v BD. imM(1;5) thuc ng thng AB v trung im E ca cnh CD thuc ng thng : x+ y 5 = 0. Vit phng trnh ng thng AB.Bi 3.7 (A-06). Trong mt phng vi h ta cac vung gc Oxy, cho ccng thng :
d1 : x+ y + 3 = 0, d2 : x y 4 = 0, d3 : x 2y = 0.Tm ta imM nm trn ng thng d3 sao cho khong cch t M n ngthng d1 bng hai ln khong cch t M n ng thng d2.
Bi 3.8 (A-05). Trong mt phng vi h ta cac vung gc Oxy, cho haing thng :
d1 : x y = 0 v d2 : 2x+ y 1 = 0.Tm ta cc nh ca hnh vung ABCD bit rng nh A thuc d1, nh Cthuc d2 v cc nh B, D thuc trc honh.
Bi 3.9 (A-04). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(0;2) v B(3;1). Tm ta trc tm v ta tm ng trn ngoitip ca tam gic OAB.
Bi 3.10 (A-02). Trong mt phng vi h ta cac vung gc Oxy, xt tamgic ABC vung ti A, phng trnh ng thng BC l
3x y 3 = 0, cc
nh A v B thuc trc honh v bn knh ng trn ni tip bng 2. Tm ta trng tm G ca tam gic ABC.
Bi 3.11 (B-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC vung ti A, c nh C(-4;1), phn gic trong gc A c phng trnhx + y 5 = 0. Vit phng trnh ng thng BC, bit din tch tam gic ABCbng 24 v nh A c honh dng.
-
Chng 3.Hnh hc gii tch trong mt phng 24
Bi 3.12 (B-09). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC cn ti A c nh A(-1;4) v cc nh B, C thuc ng thng :xy4 = 0. Xc nh ta cc im B v C, bit rng din tch tam gic ABCbng 18.
Bi 3.13 (B-08). Trong mt phng vi h ta cac vung gc Oxy, hy xcnh ta nh C ca tam gic ABC bit rng hnh chiu vung gc ca C trnng thng AB l im H(-1;-1), ng phn gic trong ca gc A c phngtrnh x y + 2 = 0 v ng cao k t B c phng trnh 4x+ 3y 1 = 0.Bi 3.14 (B-07). Trong mt phng vi h ta cac vung gc Oxy, cho imA(2;2) v cc ng thng :
d1 : x+ y 2 = 0, d2 : x+ y 8 = 0.Tm ta im B v C ln lt thuc d1 v d2 sao cho tam gic ABC vung cnti A.
Bi 3.15 (B-04). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(1;1), B(4;-3). Tm im C thuc ng thng x 2y 1 = 0 sao chokhong cch t C n ng thng AB bng 6.
Bi 3.16 (B-03). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c AB=AC, BAC = 90o. Bit M(1;-1) l trung im cnh BC v
G(2
3; 0) l trng tm tam gic ABC. Tm ta cc nh A, B, C.
Bi 3.17 (B-02). Trong mt phng vi h ta cac vung gc Oxy, cho hnhch nht ABCD c tm I(
1
2; 0), phng trnh ng thng AB l x 2y + 2 = 0
v AB=2AD. Tm ta cc nh A, B, C, D bit rng nh A c honh m.
Bi 3.18 (D-10). Trong mt phng vi h ta cac vung gc Oxy, cho imA(0;2) v l ng thng i qua O. Gi H l hnh chiu vung gc ca A trn. Vit phng trnh ng thng , bit rng khong cch t H n trc honhbng AH.
Bi 3.19 (D-09). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c M(2;0) l trung im ca cnh AB. ng trung tuyn v ng caoi qua nh A ln lt c phng trnh l 7x 2y 3 = 0 v 6x y 4 = 0. Vitphng trnh ng thng AC.
Bi 3.20 (D-04). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c cc nh A(-1;0); B(4;0); C(0;m) vi m 6= 0. Tm ta trng tmG ca tam gic ABC theo m. Xc nh m tam gic GAB vung ti G.
Chng 5.Kho st hm s 37
2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m th hm s (1) tip xc vi ng thng y = x.
Bi 5.3 (D-05). Gi (Cm) l th hm s y =1
3x3 m
2x2 +
1
3(*) (m l
tham s).1. Kho st s bin thin v v th ca hm s (*) ng vim = 2.2. Gi M l im thuc (Cm) c honh bng 1 . Tm m tip tuyn ca(Cm) ti im M song song vi ng thng 5x y = 0.
Bi 5.4 (D-07). Cho hm s y =2x
x+ 1.
1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm ta im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox, Oyti
A, B v tam gic OAB c din tch bng1
4.
Bi 5.5 (D-10). Cho hm s y = x4 x2 + 6.1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi ng
thng y =1
6x 1.
Bi 5.6 (B-04). Cho hm s y =1
3x3 2x2 + 3x (1) c th (C).
1. Kho st hm s (1).2. Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng ltip tuyn ca (C) c h s gc nh nht.
Bi 5.7 (B-06). Cho hm s y =x2 + x 1x+ 2
1. Kho st s bin thin v v th (C) ca hm s cho2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vitim cn xin ca (C).
Bi 5.8 (B-08). Cho hm s y = 4x3 6x2 + 1 (1).1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit rng tip tuyn iqua im M(1;9).
-
Chng 5
Kho st hm s
5.1 Tip tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Cc tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.3 Tng giao th . . . . . . . . . . . . . . . . . . . . . . . 405.4 Bi ton khc . . . . . . . . . . . . . . . . . . . . . . . . . . 41p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.1 Tip tuyn
Bi 5.1 (A-11). Cho hm s y =x+ 12x 1
1. Kho st s bin thin v v th (C) ca hm s cho.
2. Chng minh rng vi mi m ng thng y = x + m lun ct th (C)ti hai im phn bit A v B. Gi k1, k2 ln lt l h s gc ca cc tiptuyn vi (C) ti A v B. Tmm tng k1 + k2 t gi tr ln nht.
Bi 5.2 (D-02). Cho hm s : y =(2m 1)xm2
x 1 (1) (m l tham s).1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m= 1.
Chng 3.Hnh hc gii tch trong mt phng 25
3.2 ng trn
Bi 3.21 (D-12). Trong mt phng vi h ta Oxy, cho ng thng d : 2xy+3 = 0. Vit phng trnh ng trn c tm thuc d, ct trc Ox ti A v B, cttrc Oy ti C v D sao cho AB = CD = 2.
Bi 3.22 (B-12). Trong mt phng c h ta Oxy, cho cc ng trn (C1) :x2 + y2 =, (C2) : x2 + y2 12x + 18 = 0 v ng thng d : x y 4 = 0.Vit phng trnh ng trn c tm thuc (C2), tip xc vi d v ct (C1) ti haiim phn bit A v B sao cho AB vung gc vi d.
Bi 3.23 (D-11). Trong mt phng ta Oxy, cho im A(1; 0) v ng trn(C) : x2 + y2 2x + 4y 5 = 0. Vit phng trnh ng thng ct (C) tiim M v N sao cho tam gic AMN vung cn ti A.
Bi 3.24 (B-11). Trong mt phng ta Oxy, cho tam gic ABC c nhB(12; 1).
ng trn ni tip tam gic ABC tip xc vi cc cnh BC, CA, AB tng ngti cc im D, E, F. Cho D(3; 1) v ng thng EF c phng trnh y 3 = 0.Tm ta nh A, bit A c tung dng.
Bi 3.25 (A-11). Trong mt phng ta Oxy, cho ng thng : x+y+2 = 0v ng trn (C) : x2 +y24x2y = 0. Gi I l tm ca (C), M l im thuc. Qua M k cc tip tuyn MA v MB n (C) (A v B l cc tip im). Tmta im M, bit t gic MAIB c din tch bng 10.
Bi 3.26 (A-10). Trong mt phng vi h ta cac vung gc Oxy, cho haing thng d1 :
3x + y = 0 v d2 :
3x y = 0. Gi (T) l ng trn tip
xc vi d1 ti A, ct d2 ti hai im B v C sao cho tam gic ABC vung ti B.
Vit phng trnh ca (T), bit rng tam gic ABC c din tch bng
3
2v im
A c honh dng.
Bi 3.27 (A-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C) : x2 + y2 + 4x+ 4y + 6 = 0 v ng thng : x+my 2m+ 3 = 0,vi m l tham s thc. Gi I l tm ca ng trn (C). Tm m ct (C) tihai im phn bit A v B sao cho din tch tam gic IAB ln nht.
Bi 3.28 (A-07). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c A(0;2), B(-2;-2), v C(4;-2). Gi H l chn ng cao k t B; M vN ln lt l trung im ca cc cnh AB v BC. Vit phng trnh ng trn iqua cc im H, M, N.
-
Chng 3.Hnh hc gii tch trong mt phng 26
Bi 3.29 (B-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 2)2 + y2 = 4
5v hai ng thng 1 : x y = 0,2 : x 7y = 0.
Xc nh ta tm K v bn knh ca ng trn (C1); bit ng trn (C1) tipxc vi cc ng thng 1, 2 v tm K thuc ng trn (C).
Bi 3.30 (B-06). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): x2 + y2 2x 6y + 6 = 0 v im M(-3;1). Gi T1 v T2 l cc tipim ca cc tip tuyn k t M n (C). Vit phng trnh ng thng T1T2.
Bi 3.31 (B-05). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(2;0) v B(6;4). Vit phng trnh ng trn (C) tip xc vi trc honhti im A v khong cch t tm ca (C) n im B bng 5.
Bi 3.32 (D-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c nh A(3;-7), trc tm l H(3;-1), tm ng trn ngoi tip l I(-2;0). Xc nh ta nh C, bit C c honh dng.
Bi 3.33 (D-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + y2 = 1. Gi I l tm ca (C). Xc nh ta im M thuc(C) sao cho IMO = 30o.
Bi 3.34 (D-07). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + (y + 2)2 = 9 v ng thng d: 3x 4y +m = 0.Tm m trn d c duy nht mt im P m t c th k c hai tip tuynPA, PB ti (C) (A, B l cc tip im) sao cho tam gic PAB u.
Bi 3.35 (D-06). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): x2 + y2 2x 2y + 1 = 0 v ng thng d: x y + 3 = 0. Tm ta im M nm trn d sao cho ng trn tm M, c bn knh gp i bn knhng trn (C), tip xc ngoi vi ng trn (C).
Bi 3.36 (D-03). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + (y 2)2 = 4 v ng thng d: x y 1 = 0.1. Vit phng trnh ng trn (C) i xng vi ng trn (C) qua ng thngd.2. Tm ta cc giao im ca (C) v (C).
3.3 Cnic
Bi 3.37 (B-12). Trong mt phng vi h ta Oxy, cho hnh thoi ABCD cAC = 2BD v ng trn tip xc vi cc cnh ca hnh thoi c phng trnh
Chng 4.T hp v s phc 35
4.1 P =443
506
4.2 C13 .C412.C21 .C48 .C11 .C44 = 207900
4.3 C215.C210.C15 + C215.C110.C25 +C315.C
110.C
15 = 56875
4.4 n = 8
4.5 C412 (C25 .C14 .C13 + C15 .C24 .C13 +C15 .C
14 .C
23) = 225
4.7 k = 9
4.8 M = 34
4.10 n = 1002
4.113n+1 2n+1
n+ 1
4.12 n = 6
4.13 3516.x5
4.14 a8 = 28C812 = 126720
4.15 C610 = 210
4.16 C38 .C23 + C48 .C04 = 238
4.17 C412 = 495
4.18 n = 7, x = 4
4.19 C1011 .21 = 22
4.20 (2)4C45 + 33.C310 = 3320
4.21 C47 = 35
4.22 n = 5
4.23 z = 12i; z = 2i
4.24 |w| = 5
4.25 z1 = 2(cos 2pi3 + i sin2pi3
)z2 = 2(cos
pi3
+ i sin pi3)
4.26 |w| = 4 + 9 = 13
4.27 z = 2 i
4.28
1. z = 13i; z = 23i2. z = 2 + 2i
4.29 z = 0, z = 12 1
2i
|z| =23
4.30 Phn o z l: 2|z + iz| = 82
4.31 A = 20
4.32 x2 + (y + 1)2 = 2
4.33 z = 3 + 4i hoc z = 5
4.34 (x 3)2 + (y + 4)2 = 4
4.35 1 + i; 1 i;1 + i;1 i
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Chng 4.T hp v s phc 34
Bi 4.27 (D-11). Tm s phc z, bit :z (2 + 3i)z = 1 9i.Bi 4.28 (B-11).
1. Tm s phc z, bit: z 5 + i
3
z 1 = 0.
2. Tm phn thc v phn o ca s phc z =
(1 + i
3
1 + i
)3.
Bi 4.29 (A-11).
1. Tm tt c cc s phc z, bit z2 = |z|2 + z.2. Tnh mun ca s phc z, bit:
(2z 1)(1 + i) + (z + 1)(1 i) = 2 2i.
Bi 4.30 (A-10).1. Tm phn o ca s phc z, bit
z = (
2 + i)2(12i).
2. Cho s phc z tha mnz =
(13i)31 i . Tm mun ca s phc
z + iz.
Bi 4.31 (A-09). Gi z1 v z2 l hai nghim phc ca phng trnh z2+2z+10 =0. Tnh gi tr ca biu thc A = |z1|2 + |z2|2.Bi 4.32 (B-10). Trong mt phng vi h ta cac vung gc Oxy, tm tphp im biu din cc s phc z tha mn:
|z i| = |(1 + i)z|.
Bi 4.33 (B-09). Tm s phc z tha mn: |z (2 + i)| = 10 v zz = 25.Bi 4.34 (D-09). Trong mt phng vi h ta cac vung gc Oxy, tm tphp im biu din cc s phc z tha mn iu kin |z (3 4i)| = 2.Bi 4.35 (D-10). Tm s phc z tha mn: |z| = 2 v z2 l s thun o.
p s
Chng 3.Hnh hc gii tch trong mt phng 27
x2 + y2 = 4. Vit phng trnh chnh tc ca elip (E) i qua cc nh A, B, C, Dca hnh thoi. Bit A thuc Ox.
Bi 3.38 (A-12). Trong mt phng vi h ta Oxy, cho ng trn(C) : x2 +y2 = 8. Vit phng trnh chnh tc elip (E), bit rng (E) c di trcln bng 8 v (E) ct (C) ti bn im to thnh bn nh ca mt hnh vung.
Bi 3.39 (A-11). Trong mt phng ta Oxy, cho elip (E) :x2
4+y2
1= 1. Tm
ta cc im A v B thuc (E), c honh dng sao cho tam gic OAB cnti O v c din tch ln nht.
Bi 3.40 (A-08). Trong mt phng vi h ta cac vung gc Oxy, hy vit
phng trnh chnh tc ca elip (E) bit rng (E) c tm sai bng
5
3v hnh ch
nht c s ca (E) c chu vi bng 20.
Bi 3.41 (B-10). Trong mt phng vi h ta cac vung gc Oxy, cho im
A(2;
3) v elip (E):x2
3+y2
2= 1. Gi F1 v F2 l cc tiu im ca (E) (F1 c
honh m), M l giao im c tung dng ca ng thng AF1 vi (E), Nl im i xng ca F2 qua M. Vit phng trnh ng trn ngoi tip tam gicANF2 .
Bi 3.42 (D-08). Trong mt phng vi h ta cac vung gc Oxy, choparabol (P): y2 = 16x v im A(1;4). Hai im phn bit B,C (B v C khcA) di ng trn (P) sao cho gc BAC = 90o. Chng minh rng ng thng BClun i qua mt im c nh.
Bi 3.43 (D-02). Trong mt phng vi h ta cac vung gc Oxy, cho elip
(E) c phng trnhx2
16+y2
9= 1. Xt im M chuyn ng trn tia Ox v im N
chuyn ng trn tia Oy sao cho ng thng MN lun tip xc vi (E). Xc nhta ca M, N on MN c di nh nht. Tnh gi tr nh nht .
Bi 3.44 (D-05). Trong mt phng vi h ta cac vung gc Oxy, cho im
C(2;0) v elp (E):x2
4+y2
1= 1. Tm ta cc im A, B thuc (E), bit rng
A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.
p s
-
Chng 3.Hnh hc gii tch trong mt phng 28
3.1 A(3; 1);D(1; 3).B(1;3).C(3;1)
3.2 A(1;1);A(4; 5).
3.3 A(4; 3);C(3;1)
3.4 N(0;4),M(0;2)N(6; 2);M(6
5; 25)
3.5 B(0;4), C(4; 0)hocB(6; 2), C(2;6)
3.6 y 5 = 0; x 4y + 19 = 0
3.7 M(22;11),M(2; 1)
3.8 A(1; 1), B(0; 0), C(1;1), D(2; 0)A(1; 1), B(2; 0), C(1;1), D(0; 0)
3.9 H(
3;1), I(3; 1)
3.10 G1(7+43
3; 6+2
3
3)
G2(431
3; 62
3
3)
3.11 3x 4y + 16 = 0
3.12 B(112
; 32);C(3
2;5
2)
B(32;5
2);C(11
2; 32)
3.13 C(103
; 34)
3.14 B(1; 3), C(3; 5)B(3;1), C(5; 3)
3.15 C = (7; 3); (4311
;2711
)
3.16 B,C = (4; 0); (2;2)
3.17 A(2; 0), B(2; 2), C(3; 0), D(1;2)
3.18 (
5 1)x 2
5 2y = 0
3.19 3x 4y + 5 = 0
3.20 m = 36
3.21 (x+ 3)2 + (y + 3)2 = 10
3.22 (C) : x2 + y2 6x 6y+ 10 = 0
3.23 : y = 1; y = 3
3.24 A(3; 133
)
3.25 M(2;4) vM(3; 1)
3.26 (x+ 123)2 + (y + 2
3)2 = 1
3.27 m = 0 m = 815
3.28 x2 + y2 x+ y 2 = 0
3.29 K(85; 45);R = 2
2
5
3.30 2x+ y 3 = 0
3.31 (x 2)2 + (y 1)2 = 1(x 2)2 + (y 7)2 = 49
3.32 C(2 +65; 3)
3.33 M(32;32
)
3.34 m = 19 m = 41
3.35 M = (1; 4); (2; 1)
Chng 4.T hp v s phc 33
Bi 4.18 (A-02). Cho khai trin nh thc:(2x12 + 2
x3
)n= C0n
(2x12
)n+C1n
(2x12
)n1 (2x3
)+ +Cn1n
(2x12
)(2x3
)n1+Cnn
(2x3
)n.
(n nguyn dng). Bit rng trong khai trin C3n = 5C1n v s hng th t
bng 20n, tm n v x.
Bi 4.19 (B-07). Tm h s ca s hng cha x10 trong khai trin nh thc Niutonca (2 + x)n, bit:
3nC0n 3n1C1n + 3n2C2n 3n3C3n + + (1)nCnn = 2048(n l s nguyn dng).
Bi 4.20 (D-07). Tm h s ca x5 trong khai trin thnh a thc ca:
x(1 2x)5 + x2(1 + 3x)10.Bi 4.21 (D-04). Tm s hng khng cha x trong khai trin nh thc Niuton ca(
3x+
14x
)7vi x > 0.
Bi 4.22 (D-03). Vi n l s nguyn dng, gi a3n3 l h s ca x3n3 trongkhai trin thnh a thc ca (x2 + 1)n(x+ 2)n. Tm n a3n3 = 26n.
4.5 S phc
Bi 4.23 (D-12). Gii phng trnh z2 + 3(1 + i)z + 5i = 0 trn tp hp cc sphc.
Bi 4.24 (D-12). Cho s phc z tha mn (2 + i)z +2(1 + 2i)
1 + i= 7 + 8i. Tm
mun ca s phc w = z + 1 + i.
Bi 4.25 (B-12). Gi z1 v z2 l hai nghim phc ca phng trnh z2 2
3iz4 = 0. Vit dng lng gic ca z1 v z2
Bi 4.26 (A-12). Cho s phc z tha5(z + i)
z + 1= 2 i. Tnh mun ca s phc
w = 1 + z + z2.
-
Chng 4.T hp v s phc 32
Bi 4.10 (A-05). Tm s nguyn dng n sao cho
C12n+1 2.2C22n+1 + 3.22C32n+1 4.23C42n+1 + + (2n+ 1).22nC2n+12n+1 = 2005.Bi 4.11 (B-03). Cho n nguyn dng. Tnh tng
C0n +22 1
2C1n +
23 13
C2n + +2n+1 1n+ 1
Cnn .
Bi 4.12 (D-08). Tm s nguyn dng n tha mn h thc
C12n + C32n + + C2n12n = 2048.
4.4 H s trong khai trin nh thc
Bi 4.13 (A-12). Cho n l s nguyn dng tha mn 5Cn1n = C3n. Tm s hng
cha x5 trong khai trin nh thc Niu-tn(nx2
14 1x
)n, x 6= 0.
Bi 4.14 (A-08). Cho khai trin (1 + 2x)n = a0 + a1x + + anxn, trong n N v cc h s a0, a1, , an tha mn h thc a0 + a1
2+ + an
2n= 4096.
Tm h s ln nht trong cc s a0, a1, , an.Bi 4.15 (A-06). Tm h s ca s hng cha x26 trong khai trin nh thc Niuton
ca(
1
x4+ x7
)n, bit rng
C12n+1 + C22n+1 + + Cn2n+1 = 220 1
(n l s nguyn dng).
Bi 4.16 (A-04). Tm h s ca x8 trong khai trin thnh a thc ca [1 + x2(1x)]8.
Bi 4.17 (A-03). Tm h s ca s hng cha x8 trong khai trin nh thc Niuton
ca(
1
x3+x5)n
, bit rng
Cn+1n+4 Cnn+3 = 7(n+ 3)(n l s nguyn dng, x > 0).
Chng 3.Hnh hc gii tch trong mt phng 29
3.36 (x 3)2 + y2 = 4A(1; 0), B(3; 2)
3.37 x220
+ y2
5= 1
3.38 x216
+ y2
163
= 1
3.39 A,B = (
2;22
); (
2;22
)
3.40 x29
+ y2
4= 1
3.41 (x 1)2 + (y 23
3)2 = 4
3
3.42 I(17;4)
3.43 M(2
7; 0);N(0;
21)gtnn(MN) = 7
3.44 A,B = (27; 43
7); (2
7;4
3
7)
-
Chng 4
T hp v s phc
4.1 Bi ton m . . . . . . . . . . . . . . . . . . . . . . . . . . 304.2 Cng thc t hp . . . . . . . . . . . . . . . . . . . . . . . . 314.3 ng thc t hp khi khai trin . . . . . . . . . . . . . . . 314.4 H s trong khai trin nh thc . . . . . . . . . . . . . . . . 324.5 S phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.1 Bi ton m
Bi 4.1 (B-12). Trong mt lp hc gm c 15 hc sinh nam v 10 hc sinh n.Gio vin gi ngu nhin 4 hc sinh ln bng gii bi tp. Tnh xc sut 4 hcsinh c gi c c nam v n.
Bi 4.2 (B-05). Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3n. Hi c bao nhiu cch phn cng i thanh nin tnh nguyn v gip 3tnh mim ni, sao cho mi tnh c 4 nam v 1 n?
Bi 4.3 (B-04). Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cuhi kh, 10 cu hi trung bnh, 15 cu hi d. T 30 cu hi c th lp c
Chng 4.T hp v s phc 31
bao nhiu kim tra, mi gm 5 cu hi khc nhau, sao cho trong mi nhtthit phi 3 loi cu hi (kh, trung bnh, d) v s cu hi d khng t hn 2?
Bi 4.4 (B-02). Cho a gic u A1A2 A2n (n 2, n nguyn) ni tip ngtrn (O). Bit rng s tam gic c cc nh l 3 trong 2n im A1, A2, , A2nnhiu gp 20 ln s hnh ch nht c cc nh l 4 trong 2n imA1, A2, , A2n,tm n.
Bi 4.5 (D-06). i thanh nhin xung kch ca mt trng ph thng c 12 hcsinh, gm 5 hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Cn chn 4 hcsinh i lm nhim v, sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn.Hi c bao nhiu cch chn nh vy?
4.2 Cng thc t hp
Bi 4.6 (B-08). Cho n, k nguyn dng, k n. Chng minh rngn+ 1
n+ 2
(1
Ckn+1+
1
Ck+1n+1
)=
1
Ckn.
Bi 4.7 (B-06). Cho tp hp A gm n phn t (n 4). Bit rng, s tp con gm 4phn t ca A bng 20 ln s tp con gm 2 phn t ca A. Tm k {1, 2, , n}sao cho tp con gm k phn t ca A l ln nht.
Bi 4.8 (D-05). Tnh gi tr ca biu thcM =A4n+1 + 3A
3n
(n+ 1)!Bit rng C2n+1 + 2C
2n+2 + 2C
2n+3 + C
2n+4 = 149 (n l s nguyn dng).
4.3 ng thc t hp khi khai trin
Bi 4.9 (A-07). Chng minh rng :
1
2C12n +
1
4C32n +
1
6C52n + +
1
2nC2n12n =
22n 12n+ 1
(n l s nguyn dng).