de thi dai hoc 2002-2014
TRANSCRIPT
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TRNG THPT NGUYEN VAN TROI
TUYEN TAP
CAC E THI AI HOC
T NAM 2002 2014
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b gio dc v o to K thi tuyn sinh i hc, cao nG nm 2002 ------------------------------ Mn thi : ton chnh thc (Thi gian lm bi: 180 pht)
_____________________________________________
Cu I (H : 2,5 im; C : 3,0 im) Cho hm s : (1) ( l tham s). 23223 )1(33 mmxmmxxy +++= m1. Kho st s bin thin v v th hm s (1) khi .1=m 2. Tm k phng trnh: c ba nghim phn bit. 033 2323 =++ kkxx3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1). Cu II.(H : 1,5 im; C: 2,0 im)
Cho phng trnh : 0121loglog 2323 =++ mxx (2) ( l tham s). m
1 Gii phng trnh (2) khi .2=m 2. Tm phng trnh (2) c t nht mt nghim thuc on [m 33;1 ]. Cu III. (H : 2,0 im; C : 2,0 im )
1. Tm nghim thuc khong )2;0( ca phng trnh: .32cos2sin213sin3cossin +=
+++ x
xxxx5
2. Tnh din tch hnh phng gii hn bi cc ng: .3,|34| 2 +=+= xyxxyCu IV.( H : 2,0 im; C : 3,0 im) 1. Cho hnh chp tam gic u nh c di cnh y bng a. Gi ABCS. ,S M v ln lt N l cc trung im ca cc cnh v Tnh theo din tch tam gic , bit rng SB .SC a AMN mt phng ( vung gc vi mt phng . )AMN )(SBC 2. Trong khng gian vi h to cac vung gc Oxyz cho hai ng thng:
v .
=++=+0422042
:1 zyxzyx
+=+=+=
tztytx
2121
:2
a) Vit phng trnh mt phng cha ng thng )(P 1 v song song vi ng thng .2 b) Cho im . Tm to im )4;1;2(M H thuc ng thng 2 sao cho on thng MH c di nh nht. Cu V.( H : 2,0 im) 1. Trong mt phng vi h to cac vung gc Oxy , xt tam gic vung ti , ABC A phng trnh ng thng l BC ,033 = yx cc nh v A B thuc trc honh v bn knh ng trn ni tip bng 2. Tm ta trng tm ca tam gic . G ABC 2. Cho khai trin nh thc:
nxnn
nxxnn
xnx
n
nx
n
nxx
CCCC
+
++
+
=
+
3
1
321
13
1
21
121
0321
22222222 L ( n l s nguyn dng). Bit rng trong khai trin C v s hng th t 13 5 nn C= bng , tm v n20 n x .
----------------------------------------Ht--------------------------------------------- Ghi ch: 1) Th sinh ch thi cao ng khng lm Cu V. 2) Cn b coi thi khng gii thch g thm. H v tn th sinh:.................................................... S bo danh:.....................
1
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b gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002 chnh thc Mn thi : ton, Khi B.
(Thi gian lm bi : 180 pht)_____________________________________________
Cu I. (H : 2,0 im; C : 2,5 im) Cho hm s : ( ) 109 224 ++= xmmxy (1) (m l tham s).1. Kho st s bin thin v v th ca hm s (1) khi 1=m .2. Tm m hm s (1) c ba im cc tr.
Cu II. (H : 3,0 im; C : 3,0 im)1. Gii phng trnh: xxxx 6cos5sin4cos3sin 2222 = .2. Gii bt phng trnh: ( ) 1)729(loglog 3 xx .3. Gii h phng trnh:
++=+=.2
3
yxyxyxyx
Cu III. ( H : 1,0 im; C : 1,5 im) Tnh din tch ca hnh phng gii hn bi cc ng :
4
42xy = v
24
2xy = .
Cu IV.(H : 3,0 im ; C : 3,0 im)1. Trong mt phng vi h ta cac vung gc Oxy cho hnh ch nht ABCD c tm
0;21I , phng trnh ng thng AB l 022 =+ yx v ADAB 2= . Tm ta cc nh
DCBA ,,, bit rng nh A c honh m.2. Cho hnh lp phng 1111 DCBABCDA c cnh bng a . a) Tnh theo a khong cch gia hai ng thng BA1 v DB1 . b) Gi PNM ,, ln lt l cc trung im ca cc cnh CDBB ,1 , 11DA . Tnh gc gia hai ng thng MP v NC1 .
Cu V. (H : 1,0 im) Cho a gic u nAAA 221 L ,2( n n nguyn ) ni tip ng trn ( )O . Bit rng s tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L nhiu gp 20 ln s hnh ch nht c cc nh l 4 trong n2 im nAAA 221 ,,, L , tm n .
--------------------------------------Ht-------------------------------------------Ghi ch : 1) Th sinh ch thi cao ng khng lm Cu IV 2. b) v Cu V.
2) Cn b coi thi khng gii thch g thm.
H v tn th sinh:................................................................... S bo danh:...............................
2
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B gio dc v o to K thi Tuyn sinh i hc ,cao ng nm 2002 chnh thc Mn thi : Ton, Khi D
(Thi gian lm bi : 180 pht) _________________________________________
CuI ( H : 3 im ; C : 4 im ).
Cho hm s : ( )
1x
mx1m2y
2
= (1) ( m l tham s ).
1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m = -1.2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m th ca hm s (1) tip xc vi ng thng xy = .Cu II ( H : 2 im ; C : 3 im ).
1. Gii bt phng trnh : ( )x3x2 . 02x3x2 2 .2. Gii h phng trnh :
=++
=+
.y22
24
y4y52
x
1xx
2x3
Cu III ( H : 1 im ; C : 1 im ). Tm x thuc on [ 0 ; 14 ] nghim ng phng trnh : 04xcos3x2cos4x3cos =+ .Cu IV ( H : 2 im ; C : 2 im ).1. Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC = AD = 4 cm ;AB = 3 cm ; BC = 5 cm . Tnh khong cch t im A ti mt phng (BCD).2. Trong khng gian vi h ta cac vung gc Oxyz, cho mt phng (P) : 02yx2 =+v ng thng md :
( ) ( )( )
=++++
=+++02m4z1m2mx
01mym1x1m2 ( m l tham s ).
Xc nh m ng thng md song song vi mt phng (P).Cu V (H : 2 im ).1. Tm s nguyn dng n sao cho 243C2....C4C2C nn
n2n
1n
0n =++++ .
2. Trong mt phng vi h ta cac vung gc Oxy , cho elip (E) c phng trnh
19
y
16
x 22 =+ . Xt im M chuyn ng trn tia Ox v im N chuyn ng trn tia Oy sao chong thng MN lun tip xc vi (E). Xc nh ta ca M , N on MN c di nhnht . Tnh gi tr nh nht .
-------------------------Ht-------------------------
Ch : 1. Th sinh ch thi cao ng khng lm cu V 2. Cn b coi thi khng gii thch g thm.
H v tn th sinh : ................................................................ S bo danh.............................
3
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b gio dc v o to K thi tuyn sinh i hc, cao ng nm 2002 ------------------------------------- p n v thang im mn ton khi A
Cu Ni dung H C
I 1 23 31 xxym +==Tp xc nh Rx . )2(363' 2 =+= xxxxy ,
===20
0'2
1
xx
y
10",066" ===+= xyxyBng bin thin
+ 210x
'y +0 0
+ 0"y y + lm U 4 CT 2 C 0 li
===30
0xx
y , 4)1( =y th:
( Th sinh c th lp 2 bng bin thin)
1 ,0 0,25
0,5
0,25
1 ,5 0,5
0,5
0,5
-1 1 2 3 x0
2
4
y
4
-
I 2 Cch I. Ta c 2332323 33033 kkxxkkxx +=+=++ .t 23 3kka += Da vo th ta thy phng trnh axx =+ 23 3 c 3 nghim phn bit 43040 23 =+= ymm c 2 nghim 21 xx
v 'y i du khi qua 1x v 2x hm s t cc tr ti 1x v 2x .Ta c 23223 )1(33 mmxmmxxy +++=
( ) .2336333
1 222 mmxmmxxmx ++++
=T y ta c mmxy += 211 2 v mmxy += 222 2 .Vy phng trnh ng thng i qua 2 im cc tr l mmxy += 22 .
1 ,0 0,25
0,25
0,25
0,25
----------
0,25
0,25
0,25 0,25
1 ,0 0,25
0,25
0,25
0,25
-----------
0,25
0,25
0,25 0,25
II 1.
Vi 2=m ta c 051loglog 2323 =++ xx iu kin 0>x . t 11log23 += xt ta c 06051 22 =+=+ tttt .
23
2
1
==
tt
5,0
0,25
0,1
0,5
5
-
31 =t (loi) , 33232 33log3log2 ==== xxxt33=x tha mn iu kin 0>x .
(Th sinh c th gii trc tip hoc t n ph kiu khc)
0,25 0,5
2.
0121loglog 2323 =++ mxx (2)
iu kin 0>x . t 11log23 += xt ta c 0220121 22 =+=+ mttmtt (3)
.21log13log0]3,1[ 2333 += xtxx
Vy (2) c nghim ]3,1[ 3 khi v ch khi (3) c nghim [ ]2,1 . t tttf += 2)(Cch 1. Hm s )(tf l hm tng trn on ][ 2;1 . Ta c 2)1( =f v 6)2( =f . Phng trnh 22)(222 +=+=+ mtfmtt c nghim [ ]2;1 .20
622222
22)2(22)1(
++
++ m
mm
mfmf
Cch 2. TH1. Phng trnh (3) c 2 nghim 21 , tt tha mn 21 21
-
2.
V (0x ; )2 nn ly 31=x v
35
2=x . Ta thy 21 , xx tha mn iu
kin 212sin x . Vy cc nghim cn tm l:
31=x v
35
2=x .
(Th sinh c th s dng cc php bin i khc)
Ta thy phng trnh 3|34| 2 +=+ xxx c 2 nghim 01 =x v .52 =x Mt khc ++ 3|34| 2 xxx [ ]5;0x . Vy
( ) ( ) ( )dxxxxdxxxxdxxxxS ++++++=++= 10
3
1
225
0
2 343343|34|3
( )dxxxx +++ 53
2 343
( ) ( ) ( )dxxxdxxxdxxxS +++++= 53
23
1
21
0
2 5635
5
3
233
1
231
0
23
25
316
23
31
25
31
++
++
+= xxxxxxxS
6109
322
326
613 =++=S (.v.d.t)
(Nu th sinh v hnh th khng nht thit phi nu bt ng thc ++ 3|34| 2 xxx [ ]5;0x )
0,25
1 ,0
0,25
0,25
0,25
0,25
0,25
1 ,0
0,25
0,25
0,25
0,25
IV 1. 1 1
x510-1
y
3
32
1
8
-1
7
-
S
N
I
M C
A K
BGi K l trung im ca BC v MNSKI = . T gi thit
MNaBCMN ,22
1 == // BC I l trung im ca SK v MN .Ta c = SACSAB hai trung tuyn tng ng ANAM = AMN cn ti A MNAI .
Mt khc
( ) ( )( ) ( )
( ) ( ) SKAISBCAIMNAIAMNAI
MNAMNSBCAMNSBC
=
.
Suy ra SAK cn ti 23aAKSAA == .
2443 222222 aaaBKSBSK ===
410
843
2
222222 aaaSKSASISAAI ==
== .
Ta c 1610.
21 2aAIMNS AMN == (vdt)
ch 1) C th chng minh MNAI nh sau: ( ) ( ) AIMNSAKMNSAKBC .2) C th lm theo phng php ta : Chng hn chn h ta cac vung gc Oxyz sao cho
haSaAaCaBK ;
63;0,0;
23;0,0;0;
2,0;0;
2),0;0;0(
trong h l di ng cao SH ca hnh chp ABCS. .
0,25
0,25
0,25
0,25
0,25
0,25
0,25
0,25
8
-
2a)Cch I. Phng trnh mt phng )(P cha ng thng 1 c dng:( ) ( ) 042242 =++++ zyxzyx ( 022 + ) ( ) ( ) ( ) 044222 =+++ zyxVy ( ) 2;22; ++=Pnr .Ta c ( )2;1;12 =ur // 2 v ( ) 22 1;2;1 M( )P // ( ) ( ) ( )
=
=
PMPMunP
22
22
01;2;10. rr
Vy ( ) 02: = zxP
Cch II Ta c th chuyn phng trnh 1 sang dng tham s nh sau:
T phng trnh 1 suy ra .02 = zx t
==
==
'42'3'2
:'2 1tztytx
tx
( ) )4;3;2(,0;2;0 111 = uM r // 1 .(Ta c th tm ta im 11 M bng cch cho 020 === zyxv tnh ( )4;3;2
2121
;1211
;2212
1 =
=ur ).
Ta c ( )2;1;12 =ur // 2 . T ta c vc t php ca mt phng )(P l :[ ] ( )1;0;2, 21 == uunP rrr . Vy phng trnh mt phng )(P i qua ( )0;2;01 Mv ( )1;0;2 =Pnr l: 02 = zx .Mt khc ( ) ( ) PM 1;2;12 phng trnh mt phng cn tm l: 02 = zx
5,0 0,25
0,25 -----------
0,25
0,25
0,1 0,5
0,5 -----------
0,5
0,5
2b)b)Cch I. ( ) MHtttHH +++ 21,2,12 = ( )32;1;1 + ttt
( ) ( ) ( ) 5)1(6111263211 22222 +=+=+++= ttttttMHt gi tr nh nht khi v ch khi ( )3;3;21 Ht =Cch II. ( )tttHH 21;2;12 +++ .MH nh nht ( )4;3;210. 22 HtuMHMH == r
5,0 0,25
0,25 -----------
0,25 0,25
0,1 0,5
0,5 -----------
0,5 0,5
V 1.Ta c ( )0;1BOxBC =I . t axA = ta c );( oaA v
.33 == ayax CC Vy ( )33; aaC .T cng thc
( )( )
++=++=
CBAG
CBAG
yyyy
xxxx
3131
ta c
+
3)1(3;
312 aaG .
Cch I. Ta c : |1|2|,1|3|,1| === aBCaACaAB . Do
1
0,25
9
-
( )2123.
21 == aACABS ABC .
Ta c ( )
|1|3|1|3132 2
+=++= aaa
BCACABSr = .2
13|1| =+
a
Vy .232|1| +=a
TH1.
+++=
3326;
3347332 11 Ga
TH2
=
3326;
3134132 22 Ga .
Cch II. y C
I
O B A x
Gi I l tm ng trn ni tip ABC . V 22 == Iyr .Phng trnh ( ) 321
311.30: 0 === IxxxtgyBI .
TH1 Nu A v O khc pha i vi .321+= IxB T 2),( =ACId.3232 +=+= Ixa
++
3326;
3347
1G
TH 2. Nu A v O cng pha i vi .321= IxB Tng tta c .3212 == Ixa
3326;
3134
2G
0,25
0,25
0,25 -----------
0,25
0,25
0,25
2.T 13 5 nn CC = ta c 3n v
1
10
-
( ) ( ) 028356)2)(1(
!1!5
!3!3! 2 === nnn
nnnnn
nn
41 = n (loi) hoc .72 =nVi 7=n ta c
.4421402.2.3514022 2223
3
4
21
37 ====
xC xxx
xx
0,25
0,25
0,5
11
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002 ------------------------- p n v thang im thi chnh thc Mn ton, khi b
Cu Ni dung H C I 1 Vi 1=m ta c 108 24 += xxy l hm chn th i xng qua Oy .
Tp xc nh Rx , ( )44164' 23 == xxxxy , 0'=y
==20
xx
,34121612" 22
== xxy 320" == xy .
Bng bin thin:
+ 2320
322x
'y 0 + 0 0 + "y + 0 0 +
+ 10 + y lm U C U lm CT li CT 6 6 Hai im cc tiu : ( )6;21 A v ( )6;22 A . Mt im cc i: ( )10;0B . Hai im un:
910;
32
1U v
910;
32
2U .
Giao im ca th vi trc tung l ( )10;0B . th ct trc honh ti 4 im c honh :
64 +=x v 64 =x . (Th sinh c th lp 2 bng bin thin)
0,1
0,25
0,5
0,25
5,1
0,5
0,5
0,5
x 0
10
y
-6
-2 2
A2 A1
B
U1 U2
12
-
I 2 ( ) ( )922924' 2223 +=+= mmxxxmmxy ,
=+==
0920
0' 22 mmxx
y
Hm s c ba im cc tr phng trnh 0'=y c 3 nghim phn bit (khi 'y i du khi qua cc nghim) phng trnh 092 22 =+mmx c 2 nghim phn bit khc 0. 092 22 =+mmx
=
mmx
m
290
22 . Phng trnh 092 22 =+mmx
c 2 nghim khc 0
>
xxx
x
x
x (2).
Do 173log9 >>x nn ( ) xx 729log)1( 3 ( ) 072333729 2 xxxx (3). t xt 3= th (3) tr thnh 2938980722 xttt x . Kt hp vi iu kin (2) ta c nghim ca bt phng trnh l: 273log9 < x .
0,1
0,25
0,25
0,25
0,25
0,1
0,25
0,25
0,25
0,25
13
-
3
++=+=
).2(2)1(3
yxyxyxyx
iu kin: )3(.00
+
yxyx
( ) += == .101)1( 63 yx yxyxyx Thay yx = vo (2), gii ra ta c .1== yx Thay 1+= yx vo (2), gii ra ta c:
21,
23 == yx .
Kt hp vi iu kin (3) h phng trnh c 2 nghim:
1,1 == yx v 21,
23 == yx
Ch : Th sinh c th nng hai v ca (1) ln lu tha bc 6 di n kt qu:
+==
.1yxyx
0,1 0,25
0,25
0,25
0,25
0,1 0,25
0,25
0,25
0,25
III
Tm giao im ca hai ng cong 4
42xy = v
24
2xy = :
44
2x =24
2x 8804432
224
===+ xxxx .
Trn [ ]8;8 ta c 24
2x4
42x v do hnh i xng qua trc tung
nn dxxxS
=8
0
22
24442 21
8
0
28
0
2
22116 SSdxxdxx == .
tnh 1S ta dng php i bin tx sin4= , khi 40 t th 80 x .
tdtdx cos4= v
>4;00cos tt . Do
0,1
0,25
0,25
5,1
0,5
0,25
x 0-4 4
2
y
-2 2 2 2
2 A2 A1
4
x4y
2
=24
xy
2
=
14
-
( ) 422cos18cos1616 40
4
0
28
0
21 +=+===
dtttdtdxxS .
38
261
221
8
0
38
0
22 === xdxxS . Vy 34221 +== SSS .
Ch : Th sinh c th tnh din tch dxxxS
=
8
8
22
2444 .
0,25
0,25
0,5
0,25
IV 1
Khong cch t I n ng thng AB bng 25 5= AD v
25== IBIA .
Do BA, l cc giao im ca ng thng AB vi ng trn tm I v bn
knh 25=R . Vy ta BA, l nghim ca h :
=+
=+
22
2
25
21
022
yx
yx
Gii h ta c ( ) ( )2;2,0;2 BA (v 0
-
IV 2a) Tm khong cch gia BA1 v DB1 .
Cch I. Chn h ta cac vung gc Oxyz sao cho
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )aaDaaaCaaBaaCaAaDaBA ;;0,;;;;0;;0;;;0;0,0;;0,0;0;,0;0;0 1111 ( ) ( ) ( )0;0;,;;,;0; 1111 aBAaaaDBaaBA === v [ ] ( )22211 ;2;, aaaDBBA = .
Vy ( ) [ ][ ] 66,.,
,2
3
11
1111
11a
aa
DBBA
BADBBADBBAd === .
Cch II. ( ) DBBADCABBAADBAABBA
111111
11
.
Tng t DBCA 111 ( )111 BCADB . Gi ( )111 BCADBG = . Do aCBBBAB === 11111 nn
GGCGBGA == 11 l tm tam gic u 11BCA c cnh bng 2a . Gi I l trung im ca BA1 th IG l ng vung gc chung ca BA1 v
DB1 , nn ( ) 623
31
31, 1111
aBAICIGDBBAd ==== .
Ch :
Th sinh c th vit phng trnh mt phng ( )P cha BA1 v song song vi DB1 l: 02 =++ azyx v tnh khong cch t 1B (hoc t D ) ti ( )P ,
hoc vit phng trnh mt phng ( )Q cha DB1 v song song vi BA1 l: 022 =++ azyx v tnh khong cch t 1A (hoc t B) ti ( )Q .
0,1
0,25
0,25
0,25
0,25
0,25 0,25
0,25
5,1
0,25
0,5 0,25
0,5
0,25
0,5
0,5
x
D1
D
C1B1
A1
z
y
x
A
CB
I
G
16
-
2b) Cch I.
T Cch I ca 2a) ta tm c
aaPaaNaaM ;2;0,0;;
2,
2;0;
0.;0;2
,2;2; 11 =
=
= NCMPaaNCaaaMP .
Vy NCMP 1 .
Cch II. Gi E l trung im ca 1CC th ( ) 11CCDDME hnh chiu vung gc ca MP trn ( )11CCDD l 1ED . Ta c
NCEDNCDNCCEDCECDCNC 11110
111111 90 === . T y theo nh l ba ng vung gc ta c NCMP 1 .
0,1
0,25
0,5 0,25
0,25
0,25
0,25 0,25
V
S tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L l 32nC .
Gi ng cho ca a gic u nAAA 221 L i qua tm ng trn ( )O l ng cho ln th a gic cho c n ng cho ln.
Mi hnh ch nht c cc nh l 4 trong n2 im nAAA 221 ,,, L c cc ng cho l hai ng cho ln. Ngc li, vi mi cp ng cho ln ta c cc u mt ca chng l 4 nh ca mt hnh ch nht. Vy s hnh ch nht ni trn bng s cp ng cho ln ca a gic nAAA 221 L tc 2nC .
Theo gi thit th:
0,1
0,25
0,25
D1A1
B1 C1
C
B
A
M E
N
P
y
x
z
17
-
( )( ) ( )
( )( ) ( )2120
62212.2
!2!2!20
!32!3!220 232
===nnnnn
nn
nnCC nn
81512 == nn . Ch :
Th sinh c th tm s hnh ch nht bng cc cch khc. Nu l lun ng i
n kt qu s hnh ch nht l 2)1( nn
th cho im ti a phn ny.
0,5
18
-
B gio dc v o to K thi tuyn sinh i hc , cao ng nm 2002 Mn Ton, khi D
p n v thang im thi chnh thc
Cu Ni dung im H C
I 3 4 1. 1 1,5 Khi m = -1 ,ta c
1x
43
1x
1x3y =
= -TX : 1x - CBT : ( ) >= 1x,01x
4y
2, hm s khng c cc tr.
1/4 1/4 3ylim
x=
; =+=
+ 1x1xylim;ylim .
- BBT :
x - 1 + y/ + + + y -3 -3 - 1/4 1/4
- TC: x=1 l tim cn ng v = ylim1x . y=-3 l tim cn ngang v 3ylim
x= 1/4 1/4
- Giao vi cc trc : x = 0 y = 1; y = 0 x = - 1/3. 1/4 - th :
x
y
1/4 1/2
19
-
2. 1 1,5 Din tch cn tnh l :
dx1x
1x3S
0
3/1
=
1/4 1/2
=
0
3/1
0
3/1 1x
dx4dx3
1/4 1/4
3/1
01xln4
3
1.3 = 1/4 1/2
3
4ln41+= ( vdt).
1/4 1/4 3. 1 1
K hiu ( )
1x
mx1m2)x(f
2
= . Yu cu bi ton tng ng vi tm
m h phng trnh sau c nghim:
(H) ( )
==
.x)x(f
x)x(f//
1/4 1/4
Ta c (H)
( )( )
=
=
0
1x
mx
01x
mx
/2
2
1/4 1/4
( )( )( ) ( )
( )
=+
=
0
1x
mx1xmx2
01x
mx
2
2
2
1/4 1/4 Ta thy vi 1m ; x = m lun tho mn h ( H ) . V vy 1m , (H)
lun c nghim , ng thi khi m = 1 th h ( H ) v nghim. Do th hm s (1) tip xc vi ng thng y = x khi v ch khi 1m . S : 1m .
1/4 1/4 II 2 3 1. 1 1,5
Bt phng trnh
>=
0x3x
02x3x2
02x3x2
2
2
2
1/4 1/2
TH 1: .2
1x2x02x3x202x3x2 22 ====
1/4 1/4
TH 2:
>
>
0x3x
02x3x2
0x3x
02x3x22
2
2
2
>
-
3x2
1x <
1/4 1/4
T hai trng hp trn suy ra S: 3x2x2
1x =
1/4 1/4 2. 1 1,5
H phng trnh
==
y2
y4y52x
2x3
1/4 1/2
=+>=
0y4y5y
0y223
x
1/4 1/4
===>=
4y1y0y
0y2x
1/4 1/4
==
==
4y
2x
1y
0x
1/4 1/2
III
1 1 Phng trnh ( ) ( ) 01x2cos4xcos3x3cos =++
0xcos8xcos4 23 = ( ) 02xcosxcos4 2 = 0xcos = 1/4 1/2
+= k
2x .
1/4 1/4 [ ] 3k2k1k0k14;0x ==== 1/4
S : ;2
x=
2
3x
= ; 2
5x
= ; 2
7x
= . 1/4 1/4
IV 2 2 1. 1 1 Cch 1
T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4 Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD i
mt vung gc vi nhau.
1/4
1/4 Do c th chn h to cac vung gc, gc A sao cho B(3;0;0) ,
C(0;4;0), D( 0;0;4). Mt phng (BCD) c phng trnh :
014
z
4
y
3
x =++ . 1/4 1/4
Khong cch cn tnh l : 17
346
16
1
16
1
9
1
1 =++
(cm).
1/4 1/4
21
-
Cch 2
T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4
Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD i mt vung gc vi nhau.
1/4
1/4
D H C A E B Gi AE l ng cao ca tam gic ABC; AH l ng cao ca tam gic ADE th AH chnh l khong cch cn tnh.
D dng chng minh c h thc: 2222 AC
1
AB
1
AD
1
AH
1 ++= . 1/4 1/4
Thay AC=AD=4 cm; AB = 3 cm vo h thc trn ta tnh c:
cm17
346AH =
1/4 1/4 Cch 3:
T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4 Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD i
mt vung gc vi nhau.
1/4
1/4
Gi V l th tch t din ABCD, ta c V= 8ADACAB6
1 = .
p dng cng thc )BCD(dt
V3AH = vi V = 8 v dt(BCD) =2 34
ta tnh c cm17
346AH = .
1/2 1/2 2 1 1 Cch 1:
Mt phng (P) c vect php tuyn ( )0;1;2n . ng thng md c vec t ch phng ( )( ) ( ) ( )( )m1m;1m2; 1m2m1u 2 ++ . 1/4 1/4
Suy ra
u .
n =3(2m+1).
md song song vi (P)
)P(d
nu
m
1/4 1/4
22
-
( )
=
PA,dA
0n.u
m
Ta c : iu kin 0n.u = 2
1m =
1/4 1/4
Mt khc khi m = - 1/2 th md c phng trnh :
==
0x
01y , mi im
A( 0;1;a) ca ng thng ny u khng nm trong (P), nn iu kin ( )PA,dA m c tho mn. S : m = - 1/2 1/4 1/4 Cch 2:
Vit phng trnh dm di dng tham s ta c
=+=
+=
m)t.m(12z
t1)(2m1 y
1)tm)(2m(1 x2
1/4 1/4
md // (P) h phng trnh n t sau
=+=+=
+=
02yx2
t)m1(m2z
t)1m2(1y
t)1m2)(m1(x2
v nghim
1/4 1/4 phng trnh n t sau 3(2m+1)t+1 = 0 v nghim 1/4 1/4 m=-1/2 1/4 1/4 Cch 3:
md // (P) h phng trnh n x, y, z sau
(H) ( ) ( )
=++++=+++
=+
02m4z)1m2(mx
01myx1x1m2
02yx2
v nghim 1/4 1/4
T 2 phng trnh u ca h phng trnh trn suy ra
+=
=
3
4m2y
3
1mx
1/4 1/4 Th x , y tm c vo phng trnh th ba ta c :
)6m11m(3
1z)1m2( 2 ++=+
1/4 1/4
H (H) v nghim 2
1m =
1/4 1/4 V 2 1. 1
Ta c : ( ) =
=+n
0k
kkn
n xC1x , 1/4
Cho x = 2 ta c
==
n
0k
kkn
n 2C3 1/4
5n32433 5n === . 1/2
23
-
2. 1 Cch 1
Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trn hai tia Ox v Oy.
ng thng MN c phng trnh : 01n
y
m
x =+ 1/4
ng thng ny tip xc vi (E) khi v ch khi :
1n
19
m
116
22
=
+
.
1/4 Theo BT Csi ta c :
( )2
2
2
2
22
22222
n
m9
m
n1625
n
9
m
16nmnmMN ++=
++=+=
499.16225 =+ 7MN 1/4
ng thc xy ra
>>=+
=
0n,0m
49nm
n
m9
m
n16
22
2
2
2
2
21n,72m == .
KL: Vi ( ) ( )21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4 Cch 2
Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trn hai tia Ox v Oy.
ng thng MN c phng trnh : 01n
y
m
x =+ 1/4
ng thng ny tip xc vi (E) khi v ch khi :
1n
19
m
116
22
=
+
.
1/4 Theo bt ng thc Bunhiacpski ta c
( ) 49n
3.n
m
4.m
n
9
m
16nmnmMN
2
2222222 =
+
++=+= .
7MN 1/4
- ng thc xy ra
>>=+
=
0n,0m
7nm
n
3:n
m
4:m
22 21n,72m == .
KL: Vi ( ) ( )21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4 Cch 3:
Phng trnh tip tuyn ti im (x0 ; y0) thuc (E) : 19
yy
16
xx 00 =+ 1/4
24
-
Suy ra to ca M v N l
0;x
16M
0
v
0y
9;0N
+
+=+=
20
2
20
220
20
20
2
20
22
y
9
x
16
9
y
16
x
y
9
x
16MN
1/4 S dng bt ng thc Csi hoc Bunhiacpski (nh cch 1 hoc cch 2)
ta c : 22 7MN 1/4
- ng thc xy ra
7
213y;
7
78x 00 == .
- Khi ( ) ( )21;0N,0;72M v GTNN (MN) = 7 1/4
-----------------------Ht----------------------
25
-
B gio dc v o to K thi tuyn sinh i hc ,cao ng nm 2002 ------------------------ --------------------------------------------- Hng dn chm thi mn ton khi D Cu I: 1. -Nu TS lm sai bc no th k t tr i s khng c im.
-Nu TS xc nh ng hm s v ch tm ng 2 tim cn th c 1/4 im. 2. Nu TS lm sai bc no th k t tr i s khng c im.
3. -Nu TS dng iu kin nghim kp th khng c im. -Nu TS khng loi gi tr m = 1 th b tr 1/4 im.
Cu II: 1. -Nu TS lm sai bc no th k t tr i s khng c im.
-Nu TS kt lun nghim sai b tr 1/4 im .
-Nu TS s dng iu kin sai:
> M l trung im cnh CC . '
a) Tnh th tch khi t din 'BDA M theo a v b .
b) Xc nh t s ab
hai mt phng v ( ' )A BD ( )MBD vung gc vi nhau.
Cu 4 ( 2 im).
1) Tm h s ca s hng cha x8 trong khai trin nh thc Niutn ca n
xx
+ 53
1 , bit rng
)3(7314 += +++ nCC nnnn
( n l s nguyn dng, x > 0, l s t hp chp k ca n phn t). knC
2) Tnh tch phn +=32
52 4xx
dxI .
Cu 5 (1 im). Cho x, y, z l ba s dng v x + y + z 1. Chng minh rng
.82 1 1 1 22
22
22 +++++
zz
yy
xx
HT
Ghi ch: Cn b coi thi khng gii thch g thm. H v tn th sinh: .. . S bo danh: .
27
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003 ----------------------- Mn thi : ton khi B chnh thc Thi gian lm bi: 180 pht
_______________________________________________ Cu 1 (2 im). Cho hm s ( l tham s). 3 23 (1)y x x m= + m 1) Tm th hm s (1) c hai im phn bit i xng vi nhau qua gc ta . m 2) Kho st s bin thin v v th hm s (1) khi m =2. Cu 2 (2 im).
1) Gii phng trnh 2otg tg 4sin 2sin 2
x x xcx
+ = .
2) Gii h phng trnh
2
2
2
2
2 3
23 .
yyx
xxy
+= + =
Cu 3 (3 im). 1) Trong mt phng vi h ta cac vung gc Ox cho tam gic c y ABC
n 0, 90 .AB AC BAC= = Bit (1; 1)M l trung im cnh BC v 2 ; 03
G l trng tm tam gic . Tm ta cc nh .
ABC , , A B C
2) Cho hnh lng tr ng c y l mt hnh thoi cnh ,
gc
. ' ' ' 'ABCD A B C D ABCD an 060BAD = . Gi M l trung im cnh v l trung im cnh ' .
Chng minh rng bn im ' NAA CC
', , , B M D N'
cng thuc mt mt phng. Hy tnh di cnh ' theo a t gic AA B MDN l hnh vung.
3) Trong khng gian vi h ta cac vung gc Ox cho hai im
v im sao cho . Tnh khong cch t trung im
yz
0)(2; 0; 0), (0; 0; 8)A B C (0; 6;AC =
I ca BC n ng thng OA . Cu 4 (2 im).
1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= +
2) Tnh tch phn
4 2
0
1 2sin1 sin 2
xI dxx
= + . Cu 5 (1 im). Cho l s nguyn dng. Tnh tng n
2 3 10 1 22 1 2 1 2 1
2 3 1
nn
n n nC C C n
+ + + + + +" nC (C l s t hp chp k ca phn t). kn n ----------------------------------Ht--------------------------------- Ghi ch: Cn b coi thi khng gii thch g thm. H v tn th sinh.. S bo danh
28
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003 ---------------------- Mn thi: ton Khi D chnh thc Thi gian lm bi: 180 pht
_______________________________________________
Cu 1 (2 im).
1) Kho st s bin thin v v th ca hm s 2 2 4 (1)
2x xyx += .
2) Tm ng thng d ym : 2 2m mx m= + ct th ca hm s (1) ti hai im phn bit.
Cu 2 (2 im).
1) Gii phng trnh 2 2 2sin tg cos 02 4 2x xx = .
2) Gii phng trnh . 2 222 2x x x x + = 3
Cu 3 (3 im). 1) Trong mt phng vi h ta cac vung gc cho ng trn Oxy
4)2()1( :)( 22 =+ yxC v ng thng : 1 0d x y = . Vit phng trnh ng trn ( i xng vi ng trn qua ng thng
Tm ta cc giao im ca v . ')C
(C( )C .d
) ( ')C2) Trong khng gian vi h ta cac vung gc Oxyz cho ng thng
3 2:
1 0.kx ky z
dkx y z
0+ + = + + = Tm ng thng vung gc vi mt phng k kd ( ) : 2 5 0P x y z + = .
3) Cho hai mt phng v vung gc vi nhau, c giao tuyn l ng thng ( )P ( )Q . Trn ly hai im vi , A B AB a= . Trong mt phng ly im , trong mt phng ( ly im sao cho ,
( )P C)Q D AC BD cng vung gc vi v . Tnh bn knh mt cu ngoi tip t din v tnh khong
cch t n mt phng AC BD
AAB== ABCD
( )BCD theo . a
Cu 4 ( 2 im).
1) Tm gi tr ln nht v gi tr nh nht ca hm s 2
1
1
xyx
+=+
trn on [ ]1; 2 .
2) Tnh tch phn 2
2
0 I x x d= x .
Cu 5 (1 im).
Vi l s nguyn dng, gi n 3 3na l h s ca 3 3nx trong khai trin thnh a thc ca ( 1 . Tm n 2 ) ( 2)nx x+ + n 3 3 26na n= .
------------------------------------------------ Ht ------------------------------------------------
Ghi ch: Cn b coi thi khng gii thch g thm. H v tn th sinh:.. . S bo danh:
29
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003 p n thang im thi chnh thc Mn thi : ton Khi A
Ni dung imCu 1. 2im1)
Khi 2 1 11 .
1 1x xm y xx x
+ = = = + Tp xc nh: \{ 1 }.R +
2
2 201 2' 1 . ' 02.( 1) ( 1)
xx xy yxx x
= += + = = =
+ [ ] == 011lim)(limx
xyxx
tim cn xin ca th l: xy = . = yx 1lim tim cn ng ca th l: 1=x . Bng bin thin:
th khng ct trc honh. th ct trc tung ti im (0; 1).
1 im 0,25 0,5 0, 25
x 0 1 2 + y 0 + + 0 + + 3 y CT C 1
y
x O 1 2
3
1
1
30
-
2)
th hm s 1
2
++=
xmxmxy ct trc honh ti 2 im phn bit c honh
dng phng trnh 2( ) 0f x mx x m= + + = c 2 nghim dng phn bit khc 1
2
0
1 4 0(1) 2 1 0
1 0, 0
m
mf m
mS Pm m
= > = + = > = >
01
12 01 22
0
m
mm
m
m
-
TH2: 3 3 4
1 11 (3)22 1 1 2 0 (4).
yxy yx xy x x x xx
= = = = + = + + + =
Ta chng minh phng trnh (4) v nghim.
Cch 1. 2 2
4 2 1 1 32 0, 2 2 2
+ + = + + + > x x x x x .
Cch 2. t 4 31( ) 2 ( ) min ( ) 04
= + + = > xf x x x f x f x fR .
Trng hp ny h v nghim. Vy nghim ca h phng trnh l:
1 5 1 5 1 5 1 5( ; ) (1;1), ; , ;2 2 2 2
x y + + =
.
0, 25
Cu 3. 3im 1) Cch 1. t AB a= . Gi H l hnh chiu vung gc ca B trn AC, suy ra BH AC, m BD (AAC) BD AC, do AC (BHD) AC DH. Vy gc phng nh din [ ], ' ,B A C D l gc nBHD . Xt 'A DC vung ti D c DH l ng cao, ta c . ' . 'DH A C CD A D=
. ''
CD A DDHA C
= . 2 23 3
a a aa
= = . Tng t, 'A BC vung ti B c BH l ng
cao v 2
3aBH = .
Mt khc:
n n2 2 22 2 2 2 2 2 22 2 . cos 2. cos3 3 3a a aa BD BH DH BH DH BHD BHD= = + = + ,
do n 1cos2
BHD = n o120BHD = . Cch 2. Ta c BD AC BD AC (nh l ba ng vung gc). Tng t, BC AC (BCD) AC . Gi H l giao im ca 'A C v ( ' )BC D nBHD l gc phng ca [ ]; ' ;B A C D . Cc tam gic vung HAB, HAD, HAC bng nhau HB = HC = HD H l tm BCD u n o120BHD = .
1 im
0, 25 0, 25 0, 25 0, 25 hoc 0, 25 0,25 0,5
A
A
B C
D
D
CB
H
I
32
-
2) a) T gi thit ta c
)2
; ;() ; ;(' 0); ; ;( baaMbaaCaaC .
Vy ( ; ; 0), (0; ; )2bBD a a BM a= =JJJG JJJJG
2, ; ; 2 2ab abBD BM a =
JJJG JJJJG.
( ) 23' ; 0; , . ' .2a bBA a b BD BM BA = =
JJJG JJJG JJJJG JJJG
Do 2
'1 , . ' 6 4BDA M
a bV BD BM BA = = JJJG JJJJG JJJG
.
b) Mt phng ( )BDM c vct php tuyn l 21 , ; ; 2 2ab abn BD BM a = =
JJG JJJG JJJJG,
mt phng ( ' )A BD c vct php tuyn l 22 , ' ( ; ; )n BD BA ab ab a = = JJG JJJG JJJG
.
Do 2 2 2 2
41 2( ) ( ' ) . 0 02 2
a b a bBDM A BD n n a a b = + = =JJG JJG 1ab
= .
2 im 0, 25 0, 25 0, 25
0, 25 0, 5 0, 5
Cu 4. 2im1)
Ta c ( )1 14 3 3 3 37( 3) 7( 3)n n n n nn n n n nC C n C C C n+ ++ + + + + = + + = + ( 2)( 3) 7( 3) 2 7.2! 14 12.
2!n n n n n+ + = + + = = =
S hng tng qut ca khai trin l ( )125 60 11
3 2 212 12.
k kkk kC x x C x
=
.
Ta c
60 1182 60 11 8 4.
2
= = =k
kx x k
Do h s ca s hng cha 8x l .495)!412(!4
!12412 ==C
2) Tnh tch phn 2 3
2 25 4
xdxIx x
=+ .
t 22
44
xdxt x dtx
= + =+
v 2 2 4.x t=
Vi 5x = th 3t = , vi 2 3x = th 4t = .
Khi 2 3 4 4
22 23 35
1 1 14 2 244
xdx dtI dtt ttx x
= = = + +
4
3
1 2 1 5ln ln .4 2 4 3
tt = = +
1 im
0, 5 0, 25 0, 25
1 im 0, 25 0, 25 0,25 0, 25
A
A
B C
D
D
C B
y
x
z
33
-
Cu 5. 1im
Vi mi ,u vG G
ta c | | | | | | (*)u v u v+ +G G G G (v ( )22 22 2 2| | 2 . | | | | 2 | | . | | | | | |u v u v u v u v u v u v+ = + + + + = +G G G G G G G G G G G G ) t ,1 ;
=
xxa
=
yyb 1 ; ,
=z
zc 1 ; .
p dng bt ng thc (*) ta c | | | | | | | | | | | | .a b c a b c a b c+ + + + + +G G G G G G G G G Vy
22 2 2 2
2 2 21 1 1 1 1 1( )P x y z x y z
x y zx y z = + + + + + + + + + +
.
Cch 1. Ta c
( ) 22 22 3 31 1 1 1 9( ) 3 3 9P x y z xyz tx y z xyz t + + + + + + = + , vi ( ) 223 10 3 9x y zt xyz t + + = < .
t 29 9 1( ) 9 '( ) 9 0, 0; ( )
9Q t t Q t t Q t
t t = + = < gim trn
10;9
1( ) 82.9
Q t Q = Vy ( ) 82.P Q t
(Du = xy ra khi 13
x y z= = = ). Cch 2.
Ta c 2 2
2 2 21 1 1 1 1 1( ) 81( ) 80( )x y z x y z x y zx y z x y z
+ + + + + = + + + + + + +
21 1 118( ) 80( ) 162 80 82.x y z x y zx y z
+ + + + + + =
Vy 82.P (Du = xy ra khi 1
3x y z= = = ).
Ghi ch: Cu ny cn c nhiu cch gii khc.
0, 25 0, 25 0, 25 0, 25 hoc 0,25 0,5
34
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003 p n thang im thi chnh thc Mn thi : ton Khi B
Ni dung imCu 1. 2im1) th hm s (1) c hai im phn bit i xng nhau qua gc ta tn ti 0 0x sao cho 0 0( ) ( )y x y x= tn ti 0 0x sao cho 3 2 3 20 0 0 03 ( ) 3( )x x m x x m + = + tn ti 0 0x sao cho 203x m=
0m > . 2) Kho st s bin thin v v th hm s khi m = 2. Khi 2m = hm s tr thnh 3 23 2.y x x= + Tp xc nh : \ .
2 0' 3 6 , ' 02.
xy x x y
x== = =
" 6 6. '' 0 1.y x y x= = = "y trit tiu v i du qua 1 (1;0)x = l im un.
Bng bin thin: th ct trc honh ti cc im (1;0), (1 3;0) v ct trc tung ti im (0;2) .
1 im
0, 25
0, 25
0,25
0,25 1 im
0,25
0,25
0,25
0,25
x 0 2 + y + 0 0 + 2 + C CT y 2
x
y
O
2
21
2
35
-
Cu 2. 2im
1) Gii phng trnh: 2cotg tg 4sin 2 (1).sin 2
x x xx
+ =
iu kin: sin 0
(*).cos 0xx
Khi (1) cos sin 24sin 2sin cos sin 2x x xx x x
+ =2 2cos sin 24sin 2
sin cos sin 2x x xx x x + =
22cos 2 4sin 2 2x x + = 22cos 2 cos 2 1 0x x = cos 2 1
1cos 232
x kx
x kx
== = +=
( )kZ .
Kt hp vi iu kin (*) ta c nghim ca (1) l ( ).3
x k k= + Z
2) Gii h phng trnh
2
2
2
2
23 (1)
23 (2).
yyxxxy
+= + =
iu kin 0, 0x y . Khi h cho tng ng vi
2 2
2 22 2
( )(3 ) 03 2
3 2.3 2
x y xy x yx y y
xy xxy x
+ + == + = += +
TH1: 2 211.3 2
x y xyxy x
= = == +
TH2: 2 23 0
3 2
xy x y
xy x
+ + = = + v nghim, v t (1) v (2) ta c , 0x y > .
Vy nghim ca h phng trnh l: 1.x y= =
1 im
0,25
0,25 0,25 0,25 1 im 0,25 0,5 0,25
Cu 3. 3im1) V G l trng tm ABC vM l trung im BC nn
3 ( 1;3)MA MG= = JJJG JJJJG (0;2)A . Phng trnh BC i qua (1; 1)M v vung gc vi
( 1,3)MA = JJJG l: 1( 1) 3( 1) 0 3 4 0 (1). x y x y + + = + + = Ta thy 10MB MC MA= = = ta ,B C tha mn
phng trnh: 2 2( 1) ( 1) 10 (2). x y + + = Gii h (1),(2) ta c ta ca ,B C l (4;0), ( 2; 2).
2) Ta c ' // 'A M NC A MCN= l hnh bnh hnh, do 'A C v MN ct nhau ti trung im I ca mi ng. Mt khc ADCB l hnh bnh hnh nn trung im I ca AC cng chnh l trung im ca BD. Vy MN v BD ct nhau ti trung im I ca mi ng nn BMDN l hnh bnh hnh. Do B, M, D, N cng thuc mt mt phng. Mt khc DM2 = DA2 + AM2 = DC2 + CN2 = DN2,
hay DM = DN. Vy hnh bnh hnh BMDN l hnh thoi. Do BMDN l hnh
1 im 0,25 0,25 0,25 0,25 1 im
0,5
G A
B
C
M .
D
A
D C
B N
M I
A B
C
36
-
vung MN = BD AC = BD AC2= BD2 = BB2 +BD2 3a2 = BB2 + a2 BB= 2a AA= 2a . 3) T (0;6;0)AC =JJJG v A(2; 0; 0) suy ra C(2; 6; 0), do I(1; 3; 4). Phng trnh mt phng () qua I v vung gc vi OA l : 1 0.x = ta giao im ca () vi OA l K(1; 0; 0). khong cch t I n OA l 2 2 2(1 1) (0 3) (0 4) 5.IK = + + =
0,5
1 im 0,25
0,25
0,25
0,25
Cu 4. 2im
1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= + Tp xc nh: [ ]2; 2 .
2' 1
4
xyx
=
,
22 2
0' 0 4 2
4
xy x x x
x x
= = = =.
Ta c ( 2) 2, ( 2) 2 2, (2) 2y y y = = = , Vy
[ 2;2]max ( 2) 2 2y y
= = v [ 2;2]min ( 2) 2y y
= = .
2) Tnh tch phn
4 2
0
1 2sin .1 sin 2
xI dxx
= +
Ta c
4 42
0 0
1 2sin cos 21 sin 2 1 sin 2
x xI dx dxx x
= =+ + . t 1 sin 2 2cos 2t x dt xdx= + = . Vi 0x = th 1,t = vi
4x = th 2t = .
Khi 2
1
21 1 1ln | | ln 2.12 2 2
dtI tt
= = =
1 im 0,25 0,25 0,25
0,25 1 im 0,25 0,25
0,25 0,25
Cu 5. 1im
Ta c 0 1 2 2(1 ) ...n n nn n n nx C C x C x C x+ = + + + + .
Suy ra ( )2 2 0 1 2 21 1
(1 ) ...n n nn n n nx dx C C x C x C x dx+ = + + + + 22 2 3 1
1 0 1 2
1 1
1 (1 ) ...1 2 3 1
nn n
n n n nx x xx C x C C C
n n
++ + = + + + + + +
2 3 1 1 10 1 22 1 2 1 2 1 3 2
2 3 1 1
n n nn
n n n nC C C Cn n
+ + + + + + + =+ +" .
0,5 0,5
37
-
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003 p n thang im thi chnh thc Mn thi : ton Khi D
Ni dung imCu 1. 2im
1) Kho st s bin thin v v th ca hm s 2 2 4
2x xyx += . 1 im
Tp xc nh :R \{ 2 }.
Ta c 2 2 4 4 .
2 2x xy xx x += = +
2
2 204 4' 1 . ' 04.( 2) ( 2)
xx xy yxx x
== = = =
[ ] 4lim lim 02x x
y xx = = tim cn xin ca th l: y x= ,
tim cn ng ca th l: 2
limx
y = 2x = . Bng bin thin:
th khng ct trc honh. th ct trc tung ti im (0; 2).
0,25
0,5
0,25
2) 1 im
ng thng ct th hm s (1) ti 2 im phn bit md
phng trnh 4 2 22
x mx mx
+ = + c hai nghim phn bit khc 2 2( 1)( 2) 4m x = c hai nghim phn bit khc 2 1 0m > 1.m >
Vy gi tr cn tm l m 1.m >
0,5
0,5
x
2
6
22 4O
y
x 0 2 4 + y + 0 0 + 2 + + y C CT 6
38
-
Cu 2. 2im
1) Gii phng trnh 2 2 2 tg cos 02 4 2x xx sin (1) = 1 im
iu kin: (*). Khi cos 0x ( )221 sin 1(1) 1 cos 1 cos2 2 2cos
xx xx
= + ( ) ( )2 21 sin sin 1 cos cosx x x = + x
( ) ( )1 sin (1 cos )(1 cos ) 1 cos (1 sin )(1 sin )x x x x x + = + + x ( )1 sin (1 cos )(sin cos ) 0x x x x + + =
2sin 1 2cos 1 2tg 1
4
x kxx x kx x k
= += = = + = = +
( )kZ .
Kt hp iu kin (*) ta c nghim ca phng trnh l: 2 4
x k
x k
= + = + ( ) . kZ
0,5 0,25 0,25
2) Gii phng trnh (1). 2 222 2x x x x + 3= 1 im
t . 2
2 0x xt t= >Khi (1) tr thnh 2
4 3 3 4 0 ( 1)( 4) 0t t t t tt
= = + = = 4t (v t ) 0>
Vy 2 22 4x x x x = = 2 1
2.= =xx
Do nghim ca phng trnh l 1
2.= =xx
0,5 0,5
Cu 3. 3im1) 1 im
T ( ) suy ra c tm v bn knh 2 2: ( 1) ( 2) 4 + =C x y ( )C (1;2)I 2.R =ng thng c vct php tuyn l nd (1; 1). = uur Do ng thng i qua
v vung gc vi d c phng trnh: (1;2)I 1 21 1x y x y 3 0 = + = .
Ta giao im ca v l nghim ca h phng trnh: H d 1 0 2
(2;1).3 0 1
x y x
Hx y y = = + = =
Gi l im i xng vi qua . Khi J (1;2)I d2 3
(3;0)2 0
J H I
J H I
x x xJ
y x x= = = =
.
V i xng vi ( qua nn c tm l v bn knh
Do c phng trnh l:
( ')C
(C
)C d ( ')C2 2
(3;0)J 2.R =') ( 3) 4 +x y = .
Ta cc giao im ca ( v l nghim ca h phng trnh: )C ( ')C2 2
2 2 22 2
1 0 1( 1) ( 2) 4 1, 03, 2.( 3) 4 2 8 6 0( 3) 4
x y y xx y x yx yx y x xx y
= = + = = = = = + = + = + =
Vy ta giao im ca v ( l v ( )C ')C (1;0)A (3;2).B
0,5 0,25 0,25
39
-
2) 1 imTa c cp vect php tuyn ca hai mt phng xc nh l kd 1 (1;3 ; 1)=
uurn k
v . Vect php tuyn ca l 2 ( ; 1;1)= uurn k ( )P (1; 1; 2)= rn .
ng thng c vect ch phng l: kd2
1 2, (3 1; 1; 1 3 ) 0 k k k r
Nn 21 1 3 1.
1 1 2k k k k = = =
Vy gi tr cn tm l
0,5
0,5
3) 1 imTa c (P) (Q) v = (P) (Q), m AC AC (Q) AC AD, hay
. Tng t, ta c BD nn BD (P), do CBD . Vy A v B A, B nm trn mt cu ng knh CD.
090=CAD 090=
V bn knh ca mt cu l: 2 21
2 2CDR BC BD= = +
2 2 21 32 2
aAB AC BD= + + = . Gi H l trung im ca BC AH BC. Do BD (P) nn BD AH AH (BCD). Vy AH l khong cch t A n mt phng (BCD) v
1 2 .2 2
aAH BC= =
0,25 0,25 0,5
Cu 4. 2im
1) Tm gi tr ln nht v gi tr nh nht ca hm s2
1
1
xyx
+=+
trn on [ ]1; 2 . 1 im
2 31' .
( 1)
xyx
=+
' 0 1y x= = . Ta c
3( 1) 0, 2, (2) .5
y(1) y y = = = Vy [ ]1;2 (1) 2max y y = = v [ ]1;2min ( 1) 0.y y = =
0,5 0,5
2) Tnh tch phn 2
2
0 I x x d= x . 1 im
Ta c 2 0 0 1x x x , suy ra 1 2
2 2
0 1( ) ( ) = + I x x dx x x dx
1 22 3 3 2
0 1
1.2 3 3 2
= + = x x x x
0,5 0,5
u n n k = = r uur uur
3 1( ) || kd P u n r r
k 1.=k
.
A B
C
D
P
Q
H
40
-
Cu 5. 1im
Cch 1: Ta c ( 2 0 2 1 2 2 2 2 41) ...n n n nn n nnnx C x C x C x C
+ = + + + + , 0 1 1 2 2 2 3 3 3( 2) 2 2 2 ... 2n n n n n nn n n n
nnx C x C x C x C x C
+ = + + + + + . D dng kim tra 1, 2= =n n khng tha mn iu kin bi ton. Vi th 3n 3 3 2 3 2 2 1.n n n n nx x x x x = = Do h s ca 3 3nx trong khai trin thnh a thc ca l 2( 1) ( 2+ +n nx x )
nC3 0 3 1 1
3 3 2 . . 2. .n n n na C C C = + .
Vy 2
3 3
52 (2 3 4)26 26 73
2
= + = = = n
nn n na n n
n
Vy l gi tr cn tm (v nguyn dng). 5=n nCch 2: Ta c
2 32
3 3 22
0 0 0 0
1 2( 1) ( 2) 1 1
1 2 2 .
n nn n n
i kn n n nn i k n i i k k k
n n n ni k i k
x x xxx
x C C x C x C xxx
= = = =
+ + = + + = =
Trong khai trin trn, lu tha ca x l 3 3n khi 2 3i k = 3k
, hay Ta ch c hai trng hp tha iu kin ny l
2 3i k+ = .0,i = = hoc i 1, 1k= = .
Nn h s ca 3 3nx l . 0 3 3 1 13 3 . .2 . .2n n n n na C C C C = +
Do 2
3 3
52 (2 3 4)26 26 73
2
= + = = = n
nn n na n n
n
Vy l gi tr cn tm (v nguyn dng). 5=n n
0,75
0,25
hoc
0,75 0,25
41
-
B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004 ------------------------------ Mn thi : Ton , Khi A chnh thc Thi gian lm bi : 180 pht, khng k thi gian pht -------------------------------------------------------------- Cu I (2 im)
Cho hm s 2x 3x 3y2(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 cho AB = 1.
Cu II (2 im)
1) Gii bt phng trnh 22(x 16) 7 xx 3 >x 3 x 3
+
.
2) Gii h phng trnh 1 44
2 2
1log (y x) log 1y
x y 25.
= + =
Cu III (3 im)
1) Trong mt phng vi h ta Oxy cho hai im ( )A 0; 2 v ( )B 3; 1 . Tm ta trc tm v ta tm ng trn ngoi tip ca tam gic OAB.
2) Trong khng gian vi h ta Oxyz cho hnh chp S.ABCD c 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.
a) Tnh gc v khong cch gia hai ng thng SA, BM. b) Gi s mt phng (ABM) ct ng thng SD ti im N. Tnh th tch khi chp S.ABMN. Cu IV (2 im)
1) Tnh tch phn I = 2
1
x dx1 x 1+ .
2) Tm h s ca x8 trong khai trin thnh a thc ca 821 x (1 x) + .
Cu V (1 im) Cho tam gic ABC khng t, tha mn iu kin cos2A + 2 2 cosB + 2 2 cosC = 3. Tnh ba gc ca tam gic ABC.
------------------------------------------------------------------------------------------------------------------------ Cn b coi thi khng gii thch g thm. H v tn th sinh............................................................................S bo danh.................................................
42
-
B gio dc v o to ------------------------
chnh thc
thi tuyn sinh i hc, cao ng nm 2004 Mn: Ton, Khi B
Thi gian lm bi: 180 pht, khng k thi gian pht -------------------------------------------
Cu I (2 im)
Cho hm s y = xxx 323
1 23 + (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 l tip tuyn ca (C)
c h s gc nh nht.
Cu II (2 im)
1) Gii phng trnh xtgxx 2)sin1(32sin5 = .
2) Tm gi tr ln nht v gi tr nh nht ca hm s xxy
2ln= trn on [1; 3e ].
Cu III (3 im) 1) Trong mt phng vi h ta Oxy cho hai im A(1; 1), B(4; 3 ). Tm im C thuc ng
thng 012 = yx sao cho khong cch t C n ng thng AB bng 6.
2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn v mt y bng ( o0 < < o90 ). Tnh tang ca gc gia hai mt phng (SAB) v (ABCD) theo . Tnh th tch khi chp S.ABCD theo a v .
3) Trong khng gian vi h ta Oxyz cho im A )4;2;4( v ng thng d:
+=
=
+=
.41
1
23
tztytx
Vit phng trnh ng thng i qua im A, ct v vung gc vi ng thng d.
Cu IV (2 im)
1) Tnh tch phn I = dxx
xxe +1
lnln31.
2) Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cu hi kh, 10 cu hi trung bnh, 15 cu hi d. T 30 cu hi c th lp c bao nhiu kim tra, mi gm 5 cu hi khc nhau, sao cho trong mi nht thit phi c 3 loi cu hi (kh, trung bnh, d) v s cu hi d khng t hn 2 ?
Cu V (1 im) Xc nh m phng trnh sau c nghim
22422 1112211 xxxxxm ++=
++ . ------------------------------------------------------------------------------------------------------------------------
Cn b coi thi khng gii thch g thm. H v tn th sinh ................................................................................................. S bo danh ...........................
43
-
B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004 ------------------------ Mn: Ton, Khi D
chnh thc Thi gian lm bi: 180 pht, khng k thi gian pht ------------------------------------------- Cu I (2 im) Cho hm s 3 2y x 3mx 9x 1= + + (1) vi m l tham s.
1) Kho st hm s (1) khi m = 2. 2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.
Cu II (2 im) 1) Gii phng trnh .sin2sin)cossin2()1cos2( xxxxx =+
2) Tm m h phng trnh sau c nghim
=+
=+
.31
1
myyxx
yx
Cu III (3 im) 1) Trong mt phng vi h ta Oxy cho tam gic ABC c cc nh );0();0;4();0;1( mCBA
vi 0m . Tm ta trng tm G ca tam gic ABC theo m. Xc nh m tam gic GAB vung ti G.
2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 111. CBAABC . Bit ),0;0;(aA 0,0),;0;(),0;1;0(),0;0;( 1 >> babaBCaB .
a) Tnh khong cch gia hai ng thng CB1 v 1AC theo .,ba b) Cho ba, thay i, nhng lun tha mn 4=+ ba . Tm ba, khong cch gia hai ng
thng CB1 v 1AC ln nht.
3) Trong khng gian vi h ta Oxyz cho ba im )1;1;1(),0;0;1(),1;0;2( CBA v mt phng (P): 02 =++ zyx . Vit phng trnh mt cu i qua ba im A, B, C v c tm thuc mt phng (P).
Cu IV (2 im)
1) Tnh tch phn I = 3
2
2 )ln( dxxx .
2) Tm cc s hng khng cha x trong khai trin nh thc Niutn ca 7
43 1
+x
x vi x > 0.
Cu V (1 im) Chng minh rng phng trnh sau c ng mt nghim 01225 = xxx .
--------------------------------------------------------------------------------------------------------------------- Cn b coi thi khng gii thch g thm. H v tn th sinh.............................................................S bo danh........................................
44
-
B gio dc v o to p n - Thang im ..................... thi tuyn sinh i hc, cao ng nm 2004 ........................................... chnh thc Mn: Ton, Khi A (p n - thang im c 4 trang)
Cu Ni dung imI 2,0 I.1 (1,0 im)
( )12332
+=
xxxy = ( )
1 1x 12 2 x 1
+
.
a) Tp xc nh: { }R \ 1 . b) S bin thin:
2x(2 x)y '2(x 1)
=
; y ' 0 x 0, x 2= = = .
0,25
yC = y(2) = 12
, yCT = y(0) = 32
.
ng thng x = 1 l tim cn ng.
ng thng 1y x 12
= + l tim cn xin.
0,25
Bng bin thin: x 0 1 2 + y' 0 + + 0
y + + 12
32
0,25
c) th:
0,25
45
-
I.2 (1,0 im) Phng trnh honh giao im ca th hm s vi ng thng y = m l :
( ) mxxx
=
+
12
332 ( ) 023322 =++ mxmx (*).
0,25
Phng trnh (*) c hai nghim phn bit khi v ch khi:
0> 24m 4m 3 0 > 3m2
> hoc 1m2
< (**) .
0,25
Vi iu kin (**), ng thng y = m ct th hm s ti hai im A, B c honh x1 , x2 l nghim ca phng trnh (*). AB = 1 121 = xx
21 2x x 1 = ( )1 2 2 1 2x x 4x x 1+ =
0,25
( ) ( ) 123432 2 = mm 1 5m2
= (tho mn (**))
0,25
II 2,0 II.1 (1,0 im) iu kin : x 4 . 0,25 Bt phng trnh cho tng ng vi bt phng trnh:
2 22(x 16) x 3 7 x 2(x 16) 10 2x + > >
0,25 + Nu x > 5 th bt phng trnh c tho mn, v v tri dng, v phi m. 0,25 + Nu 4 x 5 th hai v ca bt phng trnh khng m. Bnh phng hai v ta
c: ( ) ( )22 22 x 16 10 2x x 20x 66 0 > + < 10 34 x 10 34 < < + . Kt hp vi iu kin 4 x 5 ta c: 10 34 x 5 < . p s: x 10 34>
0,25
II.2 (1,0 im) iu kin: y > x v y > 0.
( ) 11loglog 4
4
1 = yxy ( ) 11loglog 44 = yxy
0,25
4
y xlog 1y
= 4
3yx = .
0,25
Th vo phng trnh x2 + y2 = 25 ta c:
223y y 25 y 4.
4
+ = =
0,25
So snh vi iu kin , ta c y = 4, suy ra x= 3 (tha mn y > x). Vy nghim ca h phng trnh l (3; 4).
0,25
III 3,0 III.1 (1,0 im) + ng thng qua O, vung gc vi BA( 3 ; 3)
JJJG c phng trnh 3x 3y 0+ = .
ng thng qua B, vung gc vi OA(0; 2)JJJG
c phng trnh y = 1 ( ng thng qua A, vung gc vi BO( 3 ; 1)
JJJGc phng trnh 3x y 2 0+ = )
0,25
Gii h hai (trong ba) phng trnh trn ta c trc tm H( 3 ; 1) 0,25
+ ng trung trc cnh OA c phng trnh y = 1. ng trung trc cnh OB c phng trnh 3x y 2 0+ + = . ( ng trung trc cnh AB c phng trnh 3x 3y 0+ = ).
0,25
46
-
Gii h hai (trong ba) phng trnh trn ta c tm ng trn ngoi tip tam gic OAB l ( )I 3 ; 1 .
0,25
III.2.a (1,0 im) + Ta c: ( )C 2; 0; 0 , ( )D 0; 1; 0 , ( )2;0;1M ,
( )22;0;2 =SA , ( )BM 1; 1; 2= JJJJG .
0,25
Gi l gc gia SA v BM.
Ta c: ( ) SA.BM 3cos cos SA, BM 2SA . BM = = =JJJG JJJJG
JJJG JJJJGJJJG JJJJG 30 = .
0,25
+ Ta c: ( )SA, BM 2 2; 0; 2 = JJJG JJJJG , ( )AB 2; 1; 0= JJJG . 0,25 Vy:
( ) SA, BM AB 2 6d SA, BM3SA, BM
= =
JJJG JJJJG JJJGJJJG JJJJG
0,25
III.2.b (1,0 im)
Ta c MN // AB // CD N l trung im SD
2;
2
1;0N .
0,25 ( )SA 2; 0; 2 2= JJJG , ( )2;0;1 =SM , ( )22;1;0 =SB , 1SN 0; ; 22 =
JJJG
( )SA, SM 0; 4 2; 0 = JJJG JJJG .
0,25
S.ABM
1 2 2V SA,SM SB6 3 = = JJJG JJJG JJG
0,25
S.AMN
1 2V SA,SM SN6 3 = = JJJG JJJG JJJG
S.ABMN S.ABM S.AMNV V V 2= + =
0,25
IV 2,0 IV.1 (1,0 im) 2
1
xI dx1 x 1
=
+ . t: 1= xt 12 += tx tdtdx 2= . 01 == tx , 12 == tx .
0,25
47
-
Ta c:
1 1 12 32
0 0 0
t 1 t t 2I 2t dt 2 dt 2 t t 2 dt1 t 1 t t 1
+ + = = = + + + +
0,25
I 1
3 2
0
1 12 t t 2t 2 ln t 13 2
= + +
0,25
1 1 11I 2 2 2ln 2 4ln 23 2 3
= + = .
0,25
IV.2 (1, 0 im)
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
8 2 3 42 0 1 2 2 4 3 6 4 88 8 8 8 8
5 6 7 85 10 6 12 7 14 8 168 8 8 8
1 x 1 x C C x 1 x C x 1 x C x 1 x C x 1 x
C x 1 x C x 1 x C x 1 x C x 1 x
+ = + + + + + + + +
0,25
Bc ca x trong 3 s hng u nh hn 8, bc ca x trong 4 s hng cui ln hn 8. 0,25
Vy x8 ch c trong cc s hng th t, th nm, vi h s tng ng l: 3 2 4 08 3 8 4C .C , C .C
0,25
Suy ra a8 168 70 238= + = . 0,25V 1,0
Gi 3cos22cos222cos ++= CBAM
32
cos2
cos2221cos2 2
++=
CBCBA .
0,25
Do 0
2sin >
A, 1
2cos CB nn 2 AM 2cos A 4 2 sin 4
2 + .
0,25
Mt khc tam gic ABC khng t nn 0cos A , AA coscos2 . Suy ra:
42
sin24cos2 + AAM 42
sin242
sin212 2 +
=
AA
22
sin242
sin4 2 +=AA
012
sin222
=
A. Vy 0M .
0,25
Theo gi thit: M = 0
=
=
=
2
1
2sin
12
cos
coscos2
A
CBAA
A 90B C 45
= = =
0,25
48
-
B gio dc v o to p n - Thang im ..................... thi tuyn sinh i hc, cao ng nm 2004 ........................................... chnh thc Mn: Ton, Khi B (p n - thang im c 4 trang)
Cu Ni dung imI 2,0 1 Kho st hm s (1,0 im)
3 21y x 2x 3x3
= + (1).
a) Tp xc nh: R . b) S bin thin: y' = x2 4x + 3; 3,10' === xxy .
0,25
yC = y(1) = 43
, yCT = y(3) = 0; y" = 2x 4, y'' = 0 ( ) 2x 2, y 2 3 = = . th hm s li trn khong ( ; 2), lm trn khong ( 2; + ) v c im un l
2U 2;3
.
0,25
Bng bin thin: x 1 3 + y' + 0 0 +
y 43
+
0
0,25
c) th: Giao im ca th vi cc trc Ox, Oy l cc im ( ) ( )0;0 , 3;0 .
0,25
49
-
2 Vit phng trnh tip tuyn ca (C) ti im un, ...(1,0 im)
Ti im un U22;3
, tip tuyn ca (C) c h s gc 1)2(' =y . 0,25 Tip tuyn ti im un ca th (C) c phng trnh:
2 8y 1.(x 2) y x3 3
= + = + .
0,25
H s gc tip tuyn ca th (C) ti im c honh x bng: y'(x) = x2 34 + x = 1)2( 2 x 1 y' (x) y' (2), x. 0,25
Du " =" xy ra khi v ch khi x = 2 ( l honh im un). Do tip tuyn ca th (C) ti im un c h s gc nh nht. 0,25
II 2,0 1 Gii phng trnh (1,0 im) 5sinx 2 = 3 tg2x ( 1 sinx ) (1) . iu kin: cosx 0 x k ,k Z
2
+ (*). 0,25
Khi (1)
2
2
3sin x5sin x 2 (1 sin x)1 sin x
=
02sin3sin2 2 =+ xx . 0,25
2
1sin = x hoc 2sin =x (v nghim).
0,25
+
== 262
1sin kxx hoc += 2
6
5 kx , Zk ( tho mn (*)). 0,25
2 Tm gi tr ln nht v gi tr nh nht ca hm s (1,0 im)
y = 2ln xx
2ln x(2 ln x)y ' x
= 0,25
y'= 0
3
2 3
ln x 0 x 1 [1; e ]ln x 2 x e [1; e ].
= =
= = 0.25
Khi : y(1) = 0, 2 32 34 9y(e ) , y(e )e e
= = 0,25
So snh 3 gi tr trn, ta c: 33
22 [1; e ][1; e ]
4max y khi x e , min y 0 khi x 1e
= = = = .
0,25III 3,0
1 Tm im C (1,0 im) Phng trnh ng thng AB:
41
31
=
yx 4x + 3y 7 = 0. 0,25
Gi s );( yxC . Theo gi thit ta c: 012 = yx (1).
d(C, (AB)) = 6 2 2
4x 3y 37 0 (2a)4x 3y 76
4x 3y 23 0 (2b).4 3
+ =+ = + + =+ 0,25
Gii h (1), (2a) ta c: C1( 7 ; 3). 0,25
Gii h (1), (2b) ta c: 243 27C ;11 11
. 0,25
2 Tnh gc v th tch (1,0 im)
50
-
Gi giao im ca AC v BD l O th SO (ABCD) , suy ra nSAO = .
Gi trung im ca AB l M th OM AB v ABSM Gc gia hai mt phng (SAB) v
(ABCD) l nSMO .
0,25
Tam gic OAB vung cn ti O, nn === tgaSOaOAaOM2
2
2
2,
2.
Do : n SOtgSMO 2 tgOM
= = . 0,25
2 3S.ABCD ABCD
1 1 a 2 2V S .SO a tg a tg .3 3 2 6
= = = 0,50
3 Vit phng trnh ng thng (1,0 im) ng thng d c vect ch phng )4;1;2( =v . 0,25 B d )41;1;23( tttB ++ (vi mt s thc t no ).
( )AB 1 2t;3 t; 5 4t = + +JJJG . 0,25 AB d 0. =vAB 2(1 2t) (3 t) 4( 5 4t) 0 + + + = t = 1. 0,25
AB (3; 2; 1) = JJJG Phng trnh ca 1
4
2
2
3
4:
=
+=
+ zyx . 0,25
IV 2,0 1 Tnh tch phn (1,0 im)
dxx
xxIe +=1
lnln31.
t: 2 dxt 1 3ln x t 1 3ln x 2tdt 3
x= + = + = .
x 1 t 1= = , x e t 2= = . 0,25
Ta c: ( )2 22 2 4 21 1
2 t 1 2I t dt t t dt3 3 9
= = . 0,25
25 3
1
2 1 1I t t9 5 3
= . 0,25
I = 135
116.
0,25
51
-
2 Xc nh s kim tra lp c ... (1,0 im) Mi kim tra phi c s cu d l 2 hoc 3, nn c cc trng hp sau:
c 2 cu d, 2 cu trung bnh, 1 cu kh, th s cch chn l:
23625.. 15210
215 =CCC . 0,25
c 2 cu d, 1 cu trung bnh, 2 cu kh, th s cch chn l: 10500.. 25
110
215 =CCC . 0,25
c 3 cu d, 1 cu trung bnh, 1 cu kh, th s cch chn l: 22750.. 15
110
315 =CCC . 0,25
V cc cch chn trn i mt khc nhau, nn s kim tra c th lp c l: 56875227501050023625 =++ . 0,25
V Xc nh m phng trnh c nghim 1,0 iu kin: 1 x 1. t t 2 21 x 1 x= + .
Ta c: 2 21 x 1 x t 0+ , t = 0 khi x = 0. 2 4t 2 2 1 x 2 t 2= , t = 2 khi x = 1. Tp gi tr ca t l [0; 2 ] ( t lin tc trn on [ 1; 1]). 0,25
Phng trnh cho tr thnh: m ( ) 2t 2 t t 2+ = + +
2t t 2 mt 2
+ + =
+ (*)
Xt f(t) =2t t 2t 2
+ +
+ vi 0 t 2 . Ta c f(t) lin tc trn on [0; 2 ].
Phng trnh cho c nghim x Phng trnh (*) c nghim t [0; 2 ]
]2;0[]2;0[)(max)(min tfmtf .
0,25
Ta c: f '(t) = ( )2
2t 4t 0, t 0; 2t 2
+ f(t) nghch bin trn [0; 2 ]. 0,25 Suy ra:
[0; 2 ] [0; 2 ]min f (t) f ( 2) 2 1 ; max f (t) f (0) 1= = = = .
Vy gi tr ca m cn tm l 2 1 m 1 . 0,25
52
-
B gio dc v o to p n - Thang im ..................... thi tuyn sinh i hc, cao ng nm 2004 ........................................... chnh thc Mn: Ton, Khi D (p n - thang im c 4 trang)
Cu Ni dung imI 2,0 1 Kho st hm s (1,0 im)
1962 23 ++== xxxym . a) Tp xc nh: R .
b) S bin thin: 2 2y ' 3x 12x 9 3(x 4x 3)= + = + ; y ' 0 x 1, x 3= = = . 0,25
yC = y(1) = 5 , yCT = y(3) =1. y'' = 6x 12 = 0 x = 2 y = 3. th hm s li trn khong ( ; 2), lm trn khong );2( + v c im un l
)3;2(U . 0,25 Bng bin thin:
x 1 3 + y' + 0 0 +
y 5 + 1 0,25
c) th: th hm s ct trc Oy ti im (0; 1).
0,25 2 Tm m im un ca th hm s ...(1,0 im) y = x3 3mx2 + 9x + 1 (1); y' = 3x2 6mx + 9; y'' = 6x 6m .
y"= 0 x = m y = 2m3 + 9m + 1. 0,25 y" i du t m sang dng khi i qua x = m, nn im un ca th hm s
(1) l I( m; 2m3 + 9m +1). 0,25 I thuc ng thng y = x + 1 2m3 + 9m + 1 = m + 1 0,25 2m(4 m2 ) = 0 m = 0 hoc 2=m . 0,25
53
-
II 2,0 1 Gii phng trnh (1,0 im) ( 2cosx 1) (2sinx + cosx) = sin2x sinx
( 2cosx 1) (sinx + cosx) = 0. 0,25
2cosx 1= 0 cosx =1 x k2 , k2 3
= + Z .
0,25
sinx + cosx = 0 tgx = 1 x k , k4
= + Z . 0,25
Vy phng trnh c nghim l: x k23
= + v x k , k4
= + Z . 0,25
2 Tm m h phng trnh c nghim (1,0 im)
t: u = x , v y,u 0, v 0.= H cho tr thnh: 3 3u v 1u v 1 3m
+ =+ =
(*) 0,25
u v 1uv m
+ =
= u, v l hai nghim ca phng trnh: t2 t + m = 0 (**).
0,25 H cho c nghim (x; y) H (*) c nghim u 0, v 0 Phng trnh
(**) c hai nghim t khng m. 0,25
1 4m 0
1S 1 0 0 m .4
P m 0
= = =
0,25 III 3,0
1 Tnh to trng tm G ca tam gic ABC v tm m... (1,0 im) Trng tm G ca tam gic ABC c ta :
A B C A B CG Gx x x y y y mx 1; y
3 3 3+ + + +
= = = = . Vy G(1; m3
). 0,25
Tam gic ABC vung gc ti G GA.GB 0=JJJG JJJG
. 0,25 m mGA( 2; ), GB(3; )
3 3
JJJG JJJG.
0,25
GA.GB 0=JJJG JJJG 2m6 0
9 + = m 3 6 = .
0,25 2 Tnh khong cch gia B1C v AC1,... (1,0 im) a) T gi thit suy ra:
1 1C (0; 1; b), B C (a; 1; b)= JJJJG
1 1AC ( a; 1; b), AB ( 2a;0; b)= = JJJJG JJJJG
0,25
54
-
( ) 1 1 11 1 2 2
1 1
B C, AC AB abd B C, ACa bB C, AC
= = +
JJJJG JJJJG JJJJGJJJJG JJJJG .
0,25 b) p dng bt ng thc Csi, ta c:
1 1 2 2
ab ab 1 1 a bd(B C;AC ) ab 222ab 2 2a b+
= = =+
. 0,25
Du "=" xy ra khi v ch khi a = b = 2. Vy khong cch gia B1C v AC1 ln nht bng 2 khi a = b = 2. 0,25
3 Vit phng trnh mt cu (1,0 im) I(x; y; z) l tm mt cu cn tm I (P) v IA = IB = IC .
Ta c: IA2 = (x 2)2 + y2 + ( z 1)2 ; IB2 = (x 1)2 + y2 + z2 ;
IC2 = (x 1)2 + (y 1)2 + ( z 1)2 . 0,25 Suy ra h phng trnh:
=
=
=++
22
22
02
ICIB
IBIA
zyx
=+
=+
=++
1
2
2
zyzxzyx
0,25 .0;1 === yzx 0,25 == 1IAR Phng trnh mt cu l ( x 1)2 + y2 + ( z 1)2 =1. 0,25IV 2,0
1 Tnh tch phn (1,0 im)
I = 3
2
2
ln(x x)dx . t 2 22x 1du dxu ln(x x)x x
dv dx v x
==
= =.
0,25 3 332
22 2
2x 1 1I x ln(x x) dx 3ln 6 2ln 2 2 dxx 1 x 1
= = +
0,25 ( ) 3
23ln 6 2ln 2 2x ln x 1= + . 0,25
I = 3ln6 2ln2 2 ln2 = 3ln3 2. 0,25 2 Tm s hng khng cha x... (1, 0 im)
Ta c: ( )7 k7 7 kk3 374 4k 0
1 1x C xx x
=
+ = 0,25
7 k k 28 7k7 7k k3 4 127 7
k 0 k 0C x x C x
= =
= = . 0,25
S hng khng cha x l s hng tng ng vi k (k Z, 0 k 7) tho mn:
4012
728==
kk . 0,25
S hng khng cha x cn tm l 47C 35= . 0,25
55
-
V Chng minh phng trnh c nghim duy nht 1,0 x5 x2 2x 1 = 0 (1) .
(1) x5 = ( x + 1)2 0 x 0 (x + 1) 2 1 x5 1 x 1. 0,25 Vi x 1: Xt hm s 5 2f (x) x x 2x 1= . Khi f(x) l hm s lin tc
vi mi x 1. Ta c: f(1) = 3 < 0, f(2) = 23 > 0. Suy ra f(x) = 0 c nghim thuc ( 1; 2). (2) 0,25
f '( x) = 4 4 4 45x 2x 2 (2x 2x) (2x 2) x = + + . 3 4 42x(x 1) 2(x 1) x 0, x 1= + + > . 0,25
Suy ra f(x) ng bin trn [ 1; +) (3). T (1), (2), (3) suy ra phng trnh cho c ng mt nghim. 0,25
56
-
B GIO DC V O TO ----------------------- CHNH THC
THI TUYN SINH I HC, CAO NG NM 2005 Mn: TON, khi A
Thi gian lm bi: 180 pht, khng k thi gian pht ----------------------------------------
Cu I (2 im)
Gi m(C ) l th ca hm s 1y m xx
= + (*) ( m l tham s).
1) Kho st s bin thin v v th ca hm s (*) khi 1m .4
= 2) Tm m hm s (*) c cc tr v khong cch t im cc tiu ca m(C ) n tim
cn xin ca m(C ) bng 1 .2
Cu II (2 im)
1) Gii bt phng trnh 5x 1 x 1 2x 4. > 2) Gii phng trnh 2 2cos 3x cos 2x cos x 0. =
Cu III (3 im) 1) Trong mt phng vi h ta Oxy cho hai ng thng
1d : x y 0 = v 2d : 2x y 1 0.+ = Tm ta cc nh hnh vung ABCD bit rng nh A thuc 1d , nh C thuc 2d v cc nh B, D thuc trc honh.
2) Trong khng gian vi h ta Oxyz cho ng thng x 1 y 3 z 3d :1 2 1 + = = v mt
phng (P) : 2x y 2z 9 0.+ + = a) Tm ta im I thuc d sao cho khong cch t I n mt phng (P) bng 2. b) Tm ta giao im A ca ng thng d v mt phng (P). Vit phng trnh
tham s ca ng thng nm trong mt phng (P), bit i qua A v vung gc vi d.
Cu IV (2 im)
1) Tnh tch phn 2
0
sin 2x sin xI dx.1 3cos x
+= +
2) Tm s nguyn dng n sao cho 1 2 2 3 3 4 2n 2n 12n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C (2n 1).2 C 2005
++ + + + + + + + + =L
( knC l s t hp chp k ca n phn t). Cu V (1 im)
Cho x, y, z l cc s dng tha mn 1 1 1 4.x y z+ + = Chng minh rng
1 1 1 1.2x y z x 2y z x y 2z
+ + + + + + + +
------------------------------ Ht ----------------------------- Cn b coi thi khng gii thch g thm.
H v tn th sinh ................................................. s bo danh........................................
57
-
B GIO DC V O TO ------------------------- CHNH THC
THI TUYN SINH I HC, CAO NG NM 2005Mn: TON, khi B
Thi gian lm bi: 180 pht, khng k thi gian pht --------------------------------------------------
Cu I (2 im)
Gi m(C ) l th ca hm s ( )2x m 1 x m 1y
x 1+ + + += + (*) ( m l tham s).
1) Kho st s bin thin v v th ca hm s (*) khi m 1.= 2) Chng minh rng vi m bt k, th m(C ) lun lun c im cc i, im cc tiu
v khong cch gia hai im bng 20. Cu II (2 im)
1) Gii h phng trnh ( )2 39 3x 1 2 y 1
3log 9x log y 3.
+ = =
2) Gii phng trnh 1 sin x cos x sin 2x cos 2x 0.+ + + + = Cu III (3 im)
1) Trong mt phng vi h ta Oxy cho hai im A(2;0) v B(6;4) . Vit phng trnh ng trn (C) tip xc vi trc honh ti im A v khong cch t tm ca (C) n im B bng 5.
2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 1 1 1ABC.A B C vi
1A(0; 3;0), B(4;0;0), C(0;3;0), B (4;0;4). a) Tm ta cc nh 1 1A , C . Vit phng trnh mt cu c tm l A v tip xc vi
mt phng 1 1(BCC B ). b) Gi M l trung im ca 1 1A B . Vit phng trnh mt phng (P) i qua hai im
A, M v song song vi 1BC . Mt phng (P) ct ng thng 1 1A C ti im N . Tnh di on MN.
Cu IV (2 im)
1) Tnh tch phn 2
0
s in2x cosxI dx1 cosx
= + . 2) Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi c bao nhiu
cch phn cng i thanh nin tnh nguyn v gip 3 tnh min ni, sao cho mi tnh c 4 nam v 1 n?
Cu V (1 im)
Chng minh rng vi mi x ,\ ta c: x x x
x x x12 15 20 3 4 55 4 3
+ + + + . Khi no ng thc xy ra?
--------------------------------Ht--------------------------------
Cn b coi thi khng gii thch g thm. H v tn th sinh .................................................. S bo danh ...............................
58
-
B GIO DC V O TO ----------------------- CHNH THC
THI TUYN SINH I HC, CAO NG NM 2005 Mn: TON, khi D
Thi gian lm bi: 180 pht, khng k thi gian pht -------------------------------------------
Cu I (2 im)
Gi m(C ) l th ca hm s 3 21 m 1y x x
3 2 3= + (*) ( m l tham s).
1) Kho st s bin thin v v th ca hm s (*) khi m 2.= 2) Gi M l im thuc m(C ) c honh bng 1. Tm m tip tuyn ca m(C ) ti
im M song song vi ng thng 5x y 0. = Cu II (2 im)
Gii cc phng trnh sau:
1) 2 x 2 2 x 1 x 1 4.+ + + + = 2) 4 4 3cos x sin x cos x sin 3x 0.
4 4 2 + + =
Cu III (3 im)
1) Trong mt phng vi h ta Oxy cho im ( )C 2;0 v elp ( ) 2 2x yE : 1.4 1+ = Tm
ta cc im A,B thuc ( )E , bit rng hai im A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.
2) Trong khng gian vi h ta Oxyz cho hai ng thng
1x 1 y 2 z 1d :
3 1 2 + += = v 2
x y z 2 0d :
x 3y 12 0.+ = + =
a) Chng minh rng 1d v 2d song song vi nhau. Vit phng trnh mt phng (P) cha c hai ng thng 1d v 2d .
b) Mt phng ta Oxz ct hai ng thng 1 2d , d ln lt ti cc im A, B. Tnh din tch tam gic OAB ( O l gc ta ).
Cu IV (2 im)
1) Tnh tch phn ( )2 sin x0
I e cos x cos xdx.
= + 2) Tnh gi tr ca biu thc ( )
4 3n 1 nA 3AMn 1 !+ += + , bit rng
2 2 2 2n 1 n 2 n 3 n 4C 2C 2C C 149+ + + ++ + + =
( n l s nguyn dng, knA l s chnh hp chp k ca n phn t v knC l s t hp
chp k ca n phn t). Cu V (1 im) Cho cc s dng x, y, z tha mn xyz 1.= Chng minh rng
3 3 3 3 3 31 x y 1 y z 1 z x 3 3.xy yz zx
+ + + + + ++ + Khi no ng thc xy ra?
-------------------------------Ht-------------------------------- Cn b coi thi khng gii thch g thm. H v tn th sinh.............................................. S bo danh..........................................
59
-
B GIO DC V O TO --------------------- CHNH THC
P N THANG IM THI TUYN SINH I HC, CAO NG NM 2005
---------------------------------------- Mn: TON, Khi A
(p n thang im gm 4 trang)
Cu Ni dung im I 2,0
I.1 1,0 1 1 1m y x4 4 x
= = + . a) TX: \\{0}.
b) S bin thin: 2
2 2
1 1 x 4y '4 x 4x
= = , y ' 0 x 2, x 2.= = =
0,25
yC ( ) ( )CTy 2 1, y y 2 1.= = = = ng thng x 0= l tim cn ng. ng thng
1y x4
= l tim cn xin.
0,25
c) Bng bin thin:
x 2 0 2 + y + 0 0 + y
1 + + 1
0,25
d) th
0,25
60
-
I.2 1,0
2
1y ' m , y ' 0x
= = c nghim khi v ch khi m 0> .
Nu m 0> th 1 21 1y ' 0 x , xm m= = = .
0,25
Xt du y ' x 1
m 0 1
m +
y ' + 0 || 0 + Hm s lun c cc tr vi mi m 0.>
0,25
im cc tiu ca ( )mC l 1M ;2 m .m
Tim cn xin (d) : y mx mx y 0.= =
( )2 2
m 2 m md M,d .m 1 m 1
= =+ +
0,25
( ) 22
1 m 1d M;d m 2m 1 0 m 1.2 2m 1
= = + = =+ Kt lun: m 1= .
0,25
II. 2,0
II.1 1,0
Bt phng trnh: 5x 1 x 1 2x 4 > . K: 5x 1 0x 1 0 x 2.2x 4 0
0,25
Khi bt phng trnh cho tng ng vi 5x 1 2x 4 x 1 5x 1 2x 4 x 1 2 (2x 4)(x 1) > + > + +
0,25 2 2x 2 (2x 4)(x 1) x 4x 4 2x 6x 4 + > + + > +
2x 10x 0 0 x 10. < < <
0,25
Kt hp vi iu kin ta c : 2 x 10 < l nghim ca bt phng trnh cho. 0,25 II.2 1,0
Phng trnh cho tng ng vi ( ) ( )1 cos6x cos 2x 1 cos 2x 0+ + = cos6x cos 2x 1 0 =
0,25
cos8x cos 4x 2 0 + = 22cos 4x cos 4x 3 0 + =
0,25
( )= =
cos4x 1
3cos4x loi .
2
Vy ( )= = ]cos4x 1 x k k .2
0,5
61
-
III. 3,0 III.1 1,0
V ( )1A d A t; t . V A v C i xng nhau qua BD v B,D Ox nn ( )C t; t .
0,25
V 2C d nn 2t t 1 0 t 1. = = Vy ( ) ( )A 1;1 , C 1; 1 .
0,25
Trung im ca AC l ( )I 1;0 . V I l tm ca hnh vung nn IB IA 1ID IA 1= = = =
0,25
b 1 1B Ox B(b;0) b 0,b 2D Ox D(d;0) d 0,d 2d 1 1
= = = = = =
Suy ra, ( )B 0;0 v ( )D 2;0 hoc ( )B 2;0 v ( )D 0;0 . Vy bn nh ca hnh vung l ( ) ( ) ( ) ( )A 1;1 , B 0;0 , C 1; 1 , D 2;0 , hoc ( ) ( ) ( ) ( )A 1;1 , B 2;0 , C 1; 1 , D 0;0 .
0,25
III.2a 1,0
Phng trnh ca tham s ca
x 1 td : y 3 2t
z 3 t.
= = + = +
0,25
( )I d I 1 t; 3 2t;3 t + + , ( )( ) 2t 2d I, P .3
+= 0,25
( )( ) t 4d I, P 2 1 t 3 t 2.== = =
0,25
Vy c hai im ( ) ( )1 2I 3;5;7 , I 3; 7;1 . 0,25 III.2b 1,0
V A d nn ( )A 1 t; 3 2t;3 t + + . Ta c ( )A P ( ) ( ) ( )2 1 t 3 2t 2 3 t 9 0 t 1 + + + + = = . Vy ( )A 0; 1;4 .
0,25
Mt phng ( )P c vect php tuyn ( )n 2;1; 2 .= G ng thng d c vect ch phng ( )u 1;2;1= G . V ( )P v d nn c vect ch phng ( )u n,u 5;0;5 = =
JJG G G.
0,5
Phng trnh tham s ca : x ty 1z 4 t.
= = = +
0,25
62
-
IV 2,0 IV.1 1,0
2
0
(2cos x 1)sin xI dx1 3cos x
+= + .
0,25
t
2t 1cos x3t 1 3cos x3sin xdt dx.
2 1 3cos x
== + = +
x 0 t 2, x t 1.2= = = =
0,25
( )1 22 22 1
t 1 2 2I 2 1 dt 2t 1 dt.3 3 9
= + = +
0,25
23
1
2 2t 2 16 2 34t 2 1 .9 3 9 3 3 27 = + = + + =
0,25
IV.2 1,0 Ta c ( )2n 1 0 1 2 2 3 3 2n 1 2n 12n 1 2n 1 2n 1 2n 1 2n 11 x C C x C x C x ... C x+ + ++ + + + ++ = + + + + + x . \ 0,25 o hm hai v ta c
( )( ) ( )2n 1 2 3 2 2n 1 2n2n 1 2n 1 2n 1 2n 12n 1 1 x C 2C x 3C x ... 2n 1 C x++ + + ++ + = + + + + + x . \
0,25
Thay x 2= ta c: ( )1 2 2 3 3 4 2n 2n 12n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C ... 2n 1 .2 C 2n 1.++ + + + + + + + + = +
0,25
Theo gi thit ta c 2n 1 2005 n 1002+ = = . 0,25
V 1,0
Vi a,b 0> ta c : 2 1 a b 1 1 1 14ab (a b) .a b 4ab a b 4 a b
+ + + + + Du " "= xy ra khi v ch khi a b= .
0,25
p dng kt qu trn ta c: 1 1 1 1 1 1 1 1 1 1 1 1 1 (1).
2x y z 4 2x y z 4 2x 4 y z 8 x 2y 2z + + + = + + + + +
Tng t 1 1 1 1 1 1 1 1 1 1 1 1 1 (2).
x 2y z 4 2y x z 4 2y 4 x z 8 y 2z 2x + + + = + + + + +
1 1 1 1 1 1 1 1 1 1 1 1 1 (3).
x y 2z 4 2z x y 4 2z 4 x y 8 z 2x 2y + + + = + + + + +
0,5
Vy 1 1 1 1 1 1 1 1.
2x y z x 2y z x y 2z 4 x y z + + + + = + + + + + +
Ta thy trong cc bt ng thc (1), (2), (3) th du " "= xy ra khi v ch khi x y z.= = Vy ng thc xy ra khi v ch khi 3x y z .
4= = =
0,25
-------------------------------Ht-------------------------------
63
-
B GIO DC V O TO --------------------- CHNH THC
P N THANG IM THI TUYN SINH I HC, CAO NG NM 2005
---------------------------------------- Mn: TON, Khi B
(p n thang im gm 4 trang)
Cu Ni dung im I 2,0
I.1 1,0 2x 2x 2 1m 1 y x 1 .
x 1 x 1+ += = = + ++ +
a) TX: \\{ }. 1b) S bin thin: ( ) ( )
2
2 21 x 2xy ' 1
x 1 x 1+= =+ + y ' 0 x 2, x 0., = = =
0,25
yC ( ) ( )CTy 2 2, y y 0 2.= = =1
= ng thng l tim cn ng. x = ng thng l tim cn xin. y x 1= +
0,25
Bng bin thin:
x 2 1 0 + y + 0 0 + y
2 + + 2
0,25
c) th
0,25
164
-
I.2 1,0
Ta c: 1y x m
x 1= + + + .
TX: \\{ }. 1
( )( )( )2 2x x 21y ' 1 , y ' 0 x 2, x 0.
x 1 x 1
+= = = = =+ +
0,25
Xt du y '
x 2 1 0 +y + 0 || 0 +
th ca hm s (*) lun c im cc i l ( )M 2;m 3 v im cc tiu l
. ( )N 0;m 1+
0,50
( )( ) ( ) ( )( )2 2MN 0 2 m 1 m 3 20.= + + = 0,25
II. 2,0
II.1 1,0
( )2 39 3x 1 2 y 1 (1)
3log 9x log y 3 (2)
+ = =
K: x 10 y 2. <
0,25
( ) ( )3 3 3 32 3 1 log x 3log y 3 log x log y x y. + = = = 0,25
Thay vo (1) ta c y x= ( )( )x 1 2 x 1 x 1 2 x 2 x 1 2 x 1 + = + + = ( )( )x 1 2 x 0