Download - 268 Bai Tap Boi Duong Hoc Sinh Gioi Toan 9
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PHN I: BI
1. Chng minh 7 l s v t. 2. a) Chng minh : (ac + bd)2 + (ad bc)2 = (a2 + b2)(c2 + d2) b) Chng minh bt dng thc Bunhiacpxki : (ac + bd)2 (a2 + b2)(c2 + d2) 3. Cho x + y = 2. Tm gi tr nh nht ca biu thc : S = x2 + y2. 4. a) Cho a 0, b 0. Chng minh bt ng thc Cauchy : a b ab
2+
.
b) Cho a, b, c > 0. Chng minh rng : bc ca ab a b ca b c+ + + +
c) Cho a, b > 0 v 3a + 5b = 12. Tm gi tr ln nht ca tch P = ab. 5. Cho a + b = 1. Tm gi tr nh nht ca biu thc : M = a3 + b3. 6. Cho a3 + b3 = 2. Tm gi tr ln nht ca biu thc : N = a + b. 7. Cho a, b, c l cc s dng. Chng minh : a3 + b3 + abc ab(a + b + c) 8. Tm lin h gia cc s a v b bit rng : a b a b+ > - 9. a) Chng minh bt ng thc (a + 1)2 4a b) Cho a, b, c > 0 v abc = 1. Chng minh : (a + 1)(b + 1)(c + 1) 8 10. Chng minh cc bt ng thc : a) (a + b)2 2(a2 + b2) b) (a + b + c)2 3(a2 + b2 + c2) 11. Tm cc gi tr ca x sao cho : a) | 2x 3 | = | 1 x | b) x2 4x 5 c) 2x(2x 1) 2x 1. 12. Tm cc s a, b, c, d bit rng : a2 + b2 + c2 + d2 = a(b + c + d) 13. Cho biu thc M = a2 + ab + b2 3a 3b + 2001. Vi gi tr no ca a v b th M t gi tr nh nht ? Tm gi tr nh nht . 14. Cho biu thc P = x2 + xy + y2 3(x + y) + 3. CMR gi tr nh nht ca P bng 0. 15. Chng minh rng khng c gi tr no ca x, y, z tha mn ng thc sau : x2 + 4y2 + z2 2a + 8y 6z + 15 = 0
16. Tm gi tr ln nht ca biu thc : 21
Ax 4x 9
=- +
17. So snh cc s thc sau (khng dng my tnh) : a) 7 15 v 7+ b) 17 5 1 v 45+ +
c) 23 2 19
v 273
- d) 3 2 v 2 3
18. Hy vit mt s hu t v mt s v t ln hn 2 nhng nh hn 3 19. Gii phng trnh : 2 2 23x 6x 7 5x 10x 21 5 2x x+ + + + + = - - . 20. Tm gi tr ln nht ca biu thc A = x2y vi cc iu kin x, y > 0 v 2x + xy = 4. 21. Cho
1 1 1 1S .... ...
1.1998 2.1997 k(1998 k 1) 1998 1= + + + + +
- + -.
Hy so snh S v 1998
2.1999
.
22. Chng minh rng : Nu s t nhin a khng phi l s chnh phng th a l s v t. 23. Cho cc s x v y cng du. Chng minh rng :
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a) x y
2y x+
b) 2 2
2 2
x y x y0
y x y x
+ - +
c) 4 4 2 2
4 4 2 2
x y x y x y2
y x y x y x
+ - + + +
.
24. Chng minh rng cc s sau l s v t : a) 1 2+
b) 3
mn
+ vi m, n l cc s hu t, n 0. 25. C hai s v t dng no m tng l s hu t khng ? 26. Cho cc s x v y khc 0. Chng minh rng :
2 2
2 2
x y x y4 3
y x y x
+ + +
.
27. Cho cc s x, y, z dng. Chng minh rng : 2 2 2
2 2 2
x y z x y zy z x y z x+ + + + .
28. Chng minh rng tng ca mt s hu t vi mt s v t l mt s v t. 29. Chng minh cc bt ng thc :
a) (a + b)2 2(a2 + b2) b) (a + b + c)2 3(a2 + b2 + c2) c) (a1 + a2 + .. + an)
2 n(a12 + a22 + .. + an2). 30. Cho a3 + b3 = 2. Chng minh rng a + b 2. 31. Chng minh rng : [ ] [ ] [ ]x y x y+ + .
32. Tm gi tr ln nht ca biu thc : 21
Ax 6x 17
=- +
.
33. Tm gi tr nh nht ca : x y zAy z x
= + + vi x, y, z > 0. 34. Tm gi tr nh nht ca : A = x2 + y2 bit x + y = 4. 35. Tm gi tr ln nht ca : A = xyz(x + y)(y + z)(z + x) vi x, y, z 0 ; x + y + z = 1. 36. Xt xem cc s a v b c th l s v t khng nu :
a) ab v ab
l s v t.
b) a + b v ab
l s hu t (a + b 0) c) a + b, a2 v b2 l s hu t (a + b 0)
37. Cho a, b, c > 0. Chng minh : a3 + b3 + abc ab(a + b + c) 38. Cho a, b, c, d > 0. Chng minh : a b c d 2
b c c d d a a b+ + +
+ + + +
39. Chng minh rng [ ]2x bng [ ]2 x hoc [ ]2 x 1+ 40. Cho s nguyn dng a. Xt cc s c dng : a + 15 ; a + 30 ; a + 45 ; ; a + 15n.
Chng minh rng trong cc s , tn ti hai s m hai ch s u tin l 96. 41. Tm cc gi tr ca x cc biu thc sau c ngha :
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2
2 2
1 1 1 2A= x 3 B C D E x 2x
xx 4x 5 1 x 3x 2x 1- = = = = + + -
+ - - -- -2G 3x 1 5x 3 x x 1= - - - + + +
42. a) Chng minh rng : | A + B | | A | + | B | . Du = xy ra khi no ? b) Tm gi tr nh nht ca biu thc sau : 2 2M x 4x 4 x 6x 9= + + + - + . c) Gii phng trnh : 2 2 24x 20x 25 x 8x 16 x 18x 81+ + + - + = + + 43. Gii phng trnh : 2 22x 8x 3 x 4x 5 12- - - - = . 44. Tm cc gi tr ca x cc biu thc sau c ngha :
2 2
2
1 1A x x 2 B C 2 1 9x D
1 3x x 5x 6= + + = = - - =
- - +
2 22
1 xE G x 2 H x 2x 3 3 1 x
x 42x 1 x= = + - = - - + -
-+ +
45. Gii phng trnh : 2x 3x
0x 3
-=
-
46. Tm gi tr nh nht ca biu thc : A x x= + . 47. Tm gi tr ln nht ca biu thc : B 3 x x= - +
48. So snh : a) 3 1
a 2 3 v b=2
+= + b) 5 13 4 3 v 3 1- + -
c) n 2 n 1 v n+1 n+ - + - (n l s nguyn dng) 49. Vi gi tr no ca x, biu thc sau t gi tr nh nht : 2 2A 1 1 6x 9x (3x 1)= - - + + - . 50. Tnh : a) 4 2 3 b) 11 6 2 c) 27 10 2- + -
2 2d) A m 8m 16 m 8m 16 e) B n 2 n 1 n 2 n 1= + + + - + = + - + - - (n 1) 51. Rt gn biu thc : 8 41M
45 4 41 45 4 41=
+ + -.
52. Tm cc s x, y, z tha mn ng thc : 2 2 2(2x y) (y 2) (x y z) 0- + - + + + = 53. Tm gi tr nh nht ca biu thc : 2 2P 25x 20x 4 25x 30x 9= - + + - + . 54. Gii cc phng trnh sau :
2 2 2 2 2a) x x 2 x 2 0 b) x 1 1 x c) x x x x 2 0- - - - = - + = - + + - =
4 2 2d) x x 2x 1 1 e) x 4x 4 x 4 0 g) x 2 x 3 5- - + = + + + - = - + - = - 2 2 2h) x 2x 1 x 6x 9 1 i) x 5 2 x x 25- + + - + = + + - = -
k) x 3 4 x 1 x 8 6 x 1 1 l) 8x 1 3x 5 7x 4 2x 2+ - - + + - - = + + - = + + -
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55. Cho hai s thc x v y tha mn cc iu kin : xy = 1 v x > y. CMR: 2 2x y
2 2x y+
-
.
56. Rt gn cc biu thc : a) 13 30 2 9 4 2 b) m 2 m 1 m 2 m 1
c) 2 3. 2 2 3 . 2 2 2 3 . 2 2 2 3 d) 227 30 2 123 22 2
+ + + + - + - -
+ + + + + + - + + - + +
57. Chng minh rng 6 22 32 2
+ = + .
58. Rt gn cc biu thc : ( ) ( )6 2 6 3 2 6 2 6 3 2 9 6 2 6
a) C b) D2 3
+ + + - - - + - -= = .
59. So snh :
a) 6 20 v 1+ 6 b) 17 12 2 v 2 1 c) 28 16 3 v 3 2+ + + - -
60. Cho biu thc : 2A x x 4x 4= - - + a) Tm tp xc nh ca biu thc A. b) Rt gn biu thc A.
61. Rt gn cc biu thc sau : a) 11 2 10 b) 9 2 14- -
3 11 6 2 5 2 6
c)2 6 2 5 7 2 10
+ + - +
+ + - +
62. Cho a + b + c = 0 ; a, b, c 0. Chng minh ng thc : 2 2 21 1 1 1 1 1a b c a b c+ + = + +
63. Gii bt phng trnh : 2x 16x 60 x 6- + < - . 64. Tm x sao cho : 2 2x 3 3 x- + . 65. Tm gi tr nh nht, gi tr ln nht ca A = x2 + y2 , bit rng :
x2(x2 + 2y2 3) + (y2 2)2 = 1 (1) 66. Tm x biu thc c ngha:
221 16 xa) A b) B x 8x 8
2x 1x 2x 1
-= = + - +
+- -.
67. Cho biu thc : 2 2
2 2
x x 2x x x 2xA
x x 2x x x 2x
+ - - -= -
- - + -.
a) Tm gi tr ca x biu thc A c ngha. b) Rt gn biu thc A. c) Tm gi tr ca x A < 2.
68. Tm 20 ch s thp phn u tin ca s : 0,9999....9 (20 ch s 9) 69. Tm gi tr nh nht, gi tr ln nht ca : A = | x - 2 | + | y 1 | vi | x | + | y | = 5 70. Tm gi tr nh nht ca A = x4 + y4 + z4 bit rng xy + yz + zx = 1 71. Trong hai s : n n 2 v 2 n+1+ + (n l s nguyn dng), s no ln hn ?
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72. Cho biu thc A 7 4 3 7 4 3= + + - . Tnh gi tr ca A theo hai cch. 73. Tnh : ( 2 3 5)( 2 3 5)( 2 3 5)( 2 3 5)+ + + - - + - + +
74. Chng minh cc s sau l s v t : 3 5 ; 3 2 ; 2 2 3+ - +
75. Hy so snh hai s : a 3 3 3 v b=2 2 1= - - ; 5 12 5 v2
++
76. So snh 4 7 4 7 2+ - - - v s 0.
77. Rt gn biu thc : 2 3 6 8 4Q2 3 4
+ + + +=
+ +.
78. Cho P 14 40 56 140= + + + . Hy biu din P di dng tng ca 3 cn thc bc hai 79. Tnh gi tr ca biu thc x2 + y2 bit rng : 2 2x 1 y y 1 x 1- + - = . 80. Tm gi tr nh nht v ln nht ca : A 1 x 1 x= - + + . 81. Tm gi tr ln nht ca : ( )2M a b= + vi a, b > 0 v a + b 1. 82. CMR trong cc s 2b c 2 ad ; 2c d 2 ab ; 2d a 2 bc ; 2a b 2 cd+ - + - + - + - c t nht hai s dng (a, b, c, d > 0). 83. Rt gn biu thc : N 4 6 8 3 4 2 18= + + + . 84. Cho x y z xy yz zx+ + = + + , trong x, y, z > 0. Chng minh x = y = z. 85. Cho a1, a2, , an > 0 v a1a2an = 1. Chng minh: (1 + a1)(1 + a2)(1 + an) 2n. 86. Chng minh : ( )2a b 2 2(a b) ab+ + (a, b 0). 87. Chng minh rng nu cc on thng c di a, b, c lp c thnh mt tam gic th cc on thng c di a , b , c cng lp c thnh mt tam gic.
88. Rt gn : a) 2ab b a
Ab b-
= - b) 2(x 2) 8x
B2
xx
+ -=
-.
89. Chng minh rng vi mi s thc a, ta u c : 2
2
a 22
a 1
+
+. Khi no c ng thc ?
90. Tnh : A 3 5 3 5= + + - bng hai cch.
91. So snh : a) 3 7 5 2
v 6,9 b) 13 12 v 7 65
+- -
92. Tnh : 2 3 2 3
P2 2 3 2 2 3
+ -= +
+ + - -.
93. Gii phng trnh : x 2 3 2x 5 x 2 2x 5 2 2+ + - + - - - = . 94. Chng minh rng ta lun c : n
1.3.5...(2n 1) 1P
2.4.6...2n 2n 1
-= + - .
145. Trc cn thc mu : 1 1a) b)1 2 5 x x 1+ + + +
.
146. Tnh :
a) 5 3 29 6 20 b) 6 2 5 13 48 c) 5 3 29 12 5- - - + - + - - -
147. Cho ( )( )a 3 5. 3 5 10 2= - + - . Chng minh rng a l s t nhin.
148. Cho 3 2 2 3 2 2
b17 12 2 17 12 2
- += -
- +. b c phi l s t nhin khng ?
149. Gii cc phng trnh sau : ( ) ( ) ( )( ) ( )
a) 3 1 x x 4 3 0 b) 3 1 x 2 3 1 x 3 3
5 x 5 x x 3 x 3c) 2 d) x x 5 5
5 x x 3
- - + - = - = + -
- - + - -= + - =
- + -
150. Tnh gi tr ca biu thc : M 12 5 29 25 4 21 12 5 29 25 4 21= - + + - + - -
151. Rt gn : 1 1 1 1A ...1 2 2 3 3 4 n 1 n
= + + + ++ + + - +
.
152. Cho biu thc : 1 1 1 1P ...2 3 3 4 4 5 2n 2n 1
= - + - +- - - - +
a) Rt gn P. b) P c phi l s hu t khng ? 153. Tnh :
1 1 1 1A ...
2 1 1 2 3 2 2 3 4 3 3 4 100 99 99 100= + + + +
+ + + +.
154. Chng minh : 1 1 11 ... n2 3 n
+ + + + > .
155. Cho a 17 1= - . Hy tnh gi tr ca biu thc: A = (a5 + 2a4 17a3 a2 + 18a 17)2000. 156. Chng minh : a a 1 a 2 a 3- - < - - - (a 3) 157. Chng minh : 2 1x x 0
2- + > (x 0)
158. Tm gi tr ln nht ca S x 1 y 2= - + - , bit x + y = 4.
159. Tnh gi tr ca biu thc sau vi 3 1 2a 1 2aa : A4 1 1 2a 1 1 2a
+ -= = +
+ + - -.
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160. Chng minh cc ng thc sau : ( )( ) ( )a) 4 15 10 6 4 15 2 b) 4 2 2 6 2 3 1+ - - = + = +
( )( ) ( )2c) 3 5 3 5 10 2 8 d) 7 48 3 1 e) 17 4 9 4 5 5 22
- + - = + = + - + = -
161. Chng minh cc bt ng thc sau : 5 5 5 5
a) 27 6 48 b) 10 05 5 5 5
+ -+ > + -
+ + + -
2 3 1 2 3 3 3 1d) 3 2 0
2 6 2 6 2 6 2 6 2
+ - -+ + - + - >
+ - +
e) 2 2 2 1 2 2 2 1 1,9 g) 17 12 2 2 3 1+ - + - - > + - > -
( ) ( ) 2 2 3 2 2h) 3 5 7 3 5 7 3 i) 0,84+ + -
+ + - + + < <
162. Chng minh rng : 12 n 1 2 n 2 n 2 n 1n
+ - < < - - . T suy ra:
1 1 1
2004 1 ... 20052 3 1006009
< + + + + <
163. Trc cn thc mu : 3 3
2 3 4 3a) b)
2 3 6 8 4 2 2 4
+ ++ + + + + +
.
164. Cho 3 2 3 2
x v y=3 2 3 2
+ -=
- +. Tnh A = 5x2 + 6xy + 5y2.
165. Chng minh bt ng thc sau : 2002 2003 2002 20032003 2002
+ > + .
166. Tnh gi tr ca biu thc : 2 2x 3xy y
Ax y 2- +
=+ +
vi x 3 5 v y 3 5= + = - .
167. Gii phng trnh : 26x 3 3 2 x xx 1 x
-= + -
- -.
168. Gii bt cc pt : a)
13 3 5x 72 b) 10x 14 1 c) 2 2 2 2x 4
4+ - + + .
169. Rt gn cc biu thc sau : a 1
a) A 5 3 29 12 5 b) B 1 a a(a 1) aa-
= - - - = - + - +
2 2 2
2 2 2
x 3 2 x 9 x 5x 6 x 9 xc) C d) D
2x 6 x 9 3x x (x 2) 9 x
+ + - + + + -= =
- + - - + + -
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1 1 1 1E ...
1 2 2 3 3 4 24 25= - + - -
- - - -
170. Tm GTNN v GTLN ca biu thc 2
1A
2 3 x=
- -.
171. Tm gi tr nh nht ca 2 1A1 x x
= +-
vi 0 < x < 1.
172. Tm GTLN ca : a) A x 1 y 2= - + - bit x + y = 4 ;
b) y 2x 1
Bx y
--= +
173. Cho a 1997 1996 ; b 1998 1997= - = - . So snh a vi b, s no ln hn ? 174. Tm GTNN, GTLN ca :
2
2
1a) A b) B x 2x 4
5 2 6 x= = - + +
+ -.
175. Tm gi tr ln nht ca 2A x 1 x= - . 176. Tm gi tr ln nht ca A = | x y | bit x2 + 4y2 = 1. 177. Tm GTNN, GTLN ca A = x3 + y3 bit x, y 0 ; x2 + y2 = 1. 178. Tm GTNN, GTLN ca A x x y y= + bit x y 1+ = .
179. Gii phng trnh : 2 x 11 x x 3x 2 (x 2) 3x 2-
- + - + + - =-
.
180. Gii phng trnh : 2 2x 2x 9 6 4x 2x+ - = + + . 181. CMR, "n Z+ , ta c :
1 1 1 1... 2
2 3 2 4 3 (n 1) n+ + + + 0 ; a 1)
186. Chng minh : a 1 a 1 14 a a 4aa 1 a 1 a
+ - - + - = - + . (a > 0 ; a 1)
187. Rt gn : ( )2
x 2 8x2
xx
+ -
- (0 < x < 2)
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188. Rt gn : b ab a b a ba :a b ab b ab a ab
- + + + - + + -
189. Gii bt phng trnh : ( )2
2 2
2 2
5a2 x x a
x a+ +
+ (a 0)
190. Cho ( )2 1 a a 1 a aA 1 a : a a 11 a 1 a
- += - + - +
- +
a) Rt gn biu thc A. b) Tnh gi tr ca A vi a = 9. c) Vi gi tr no ca a th | A | = A.
191. Cho biu thc : a b 1 a b b bBa ab 2 ab a ab a ab
+ - -= + +
+ - + .
a) Rt gn biu thc B. b) Tnh gi tr ca B nu a 6 2 5= + . c) So snh B vi -1.
192. Cho 1 1 a b
A : 1a a b a a b a b
+ = + + - - + + -
a) Rt gn biu thc A. b) Tm b bit | A | = -A. c) Tnh gi tr ca A khi a 5 4 2 ; b 2 6 2= + = + .
193. Cho biu thc a 1 a 1 1A 4 a aa 1 a 1 a
+ - = - + - - +
a) Rt gn biu thc A. b) Tm gi tr ca A nu 6a
2 6=
+. c) Tm gi tr ca a A A> .
194. Cho biu thc a 1 a a a aA2 2 a a 1 a 1
- += - -
+ - .
a) Rt gn biu thc A. b) Tm gi tr ca A A = - 4
195. Thc hin php tnh : 1 a 1 a 1 a 1 aA :1 a 1 a 1 a 1 a
+ - + -= + -
- + - +
196. Thc hin php tnh : 2 3 2 3B2 2 3 2 2 3
+ -= +
+ + - -
197. Rt gn cc biu thc sau :
( )3x y 1 1 1 2 1 1
a) A : . .x yxy xy x y 2 xy x yx y
- = + + + + + +
vi x 2 3 ; y 2 3= - = + .
b) 2 2 2 2x x y x x y
B2(x y)
+ - - - -=
- vi x > y > 0
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c) 2
2
2a 1 xC
1 x x
+=
+ - vi 1 1 a ax
2 a 1 a
-= -
- ; 0 < a < 1
`d) ( )( )2 2
2
a 1 b 1D (a b)
c 1
+ += + -
+ vi a, b, c > 0 v ab + bc + ca = 1
e) x 2 x 1 x 2 x 1
E . 2x 1x 2x 1 x 2x 1
+ - + - -= -
+ - + - -
198. Chng minh : 2 2x 4 x 4 2x 4
x xx x x
- - ++ + - = vi x 2.
199. Cho 1 2 1 2
a , b2 2
- + - -= = . Tnh a7 + b7.
200. Cho a 2 1= - ` a) Vit a2 ; a3 di dng m m 1- - , trong m l s t nhin.
b) Chng minh rng vi mi s nguyn dng n, s an vit c di dng trn. 201. Cho bit x = 2 l mt nghim ca phng trnh x3 + ax2 + bx + c = 0 vi cc h s hu t. Tm cc nghim cn li. 202. Chng minh 1 1 12 n 3 ... 2 n 2
2 3 n- < + + + < - vi n N ; n 2.
203. Tm phn nguyn ca s 6 6 ... 6 6+ + + + (c 100 du cn). 204. Cho 2 3a 2 3. Tnh a) a b) a = + .
205. Cho 3 s x, y, x y+ l s hu t. Chng minh rng mi s x , y u l s hu t 206. CMR, "n 1 , n N : 1 1 1 1... 2
2 3 2 4 3 (n 1) n+ + + + 0. Bit a b c d 11 a 1 b 1 c 1 d
+ + + + + + +
. Chng minh rng : 1abcd81
.
224. Chng minh bt ng thc : 2 2 2
2 2 2
x y z x y zy z x y z x+ + + + vi x, y, z > 0
225. Cho 3 33 3 3a 3 3 3 3 ; b 2 3= + + - = . Chng minh rng : a < b.
226. a) Chng minh vi mi s nguyn dng n, ta c : n
11 3
n +
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232. Gii cc phng trnh sau : 33 3a) 1 x 16 x 3 b) 2 x x 1 1+ - = + - + - =
33 3 33c) x 1 x 1 5x d) 2 2x 1 x 1+ + - = - = +
( )3 2 2 3 3333
x 3x x 1 x 4 7 x x 5e) 2 3 g) 6 x
2 7 x x 5
- - - - - - -= - = -
- + -
32 2 2 3 3 33 3h) (x 1) (x 1) x 1 1 i) x 1 x 2 x 3 0+ + - + - = + + + + + =
24 4 4 4 4 4k) 1 x 1 x 1 x 3 l) a x b x a b 2x- + + + - = - + - = + - (a, b l
tham s)
233. Rt gn 4 2 2 43 3 3
2 23 33
a a b bA
a ab b
+ +=
+ +.
234. Tm gi tr nh nht ca biu thc : 2 2A x x 1 x x 1= - + + + + 235. Xc nh cc s nguyn a, b sao cho mt trong cc nghim ca phng trnh : 3x3 + ax2 + bx + 12 = 0 l 1 3+ . 236. Chng minh 3 3 l s v t. 237. Lm php tnh : 3 6 6 3a) 1 2 . 3 2 2 b) 9 4 5. 2 5+ - + - .
238. Tnh : 3 3a 20 14 2 20 14 2= + + - .
239. Chng minh : 3 37 5 2 7 2 5 2+ + - = . 240. Tnh : ( )4 4 4A 7 48 28 16 3 . 7 48= + - - + . 241. Hy lp phng trnh f(x) = 0 vi h s nguyn c mt nghim l : 3 3x 3 9= + . 242. Tnh gi tr ca biu thc : M = x3 + 3x 14 vi 3
3
1x 7 5 2
7 5 2= + -
+.
243. Gii cc phng trnh : a) 3 3x 2 25 x 3+ + - = . 2 2 243b) x 9 (x 3) 6 c) x 32 2 x 32 3- = - + + - + =
244. Tm GTNN ca biu thc : ( ) ( )3 3 3 3A x 2 1 x 1 x 2 1 x 1= + + + + + - + . 245. Cho cc s dng a, b, c, d. Chng minh : a + b + c + d 44 abcd .
246. Rt gn : 3 32 23
3
3 3 3 3 2
8 x x 2 x x 4P : 2 x
2 x 2 x x 2 x 2 x
- -= + + + - + - +
;
Voi x > 0 , x 8 247. CMR : 3 3x 5 17 5 17= - + + l nghim ca phng trnh x3 6x 10 = 0. 248. Cho 3
3
1x 4 15
4 15= + -
-. Tnh gi tr biu thc y = x3 3x + 1987.
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249. Chng minh ng thc : 33 233 3
a 2 5. 9 4 5a 1
2 5. 9 4 5 a a
+ + -= - -
- + - +.
250. Chng minh bt ng thc : 3 3 39 4 5 2 5 . 5 2 2,1 0 + + + - -
-
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2 a a 2 a a a a 1
Da 1a 2 a 1 a
+ - + - -= -
-+ + vi a > 0 ; a 1
266. Cho biu thc c ac 1B aa c a ca c
ac c ac a ac
-= + - ++ + -
+ -
.
a) Rt gn biu thc B. b) Tnh gi tr ca biu thc B khi c = 54 ; a = 24 c) Vi gi tr no ca a v c B > 0 ; B < 0.
267. Cho biu thc : 2 2 22mn 2mn 1
A= m+ m 11+n 1 n n
+ - +
+ vi m 0 ; n 1
a) Rt gn biu thc A. b) Tm gi tr ca A vi m 56 24 5= + . c) Tm gi tr nh nht ca A. 268. Rt gn
22 2
1 x 1 x 1 1 x xD 1
x x1 x 1 x 1 x 1 x 1 x 1 x
+ - -= - - -
+ - - - - + - + -
269. Cho 1 2 x 2 x
P : 1x 1x 1 x x x x 1
= - -
+- + - - vi x 0 ; x 1.
a) Rt gn biu thc P. b) Tm x sao cho P < 0. 270. Xt biu thc
2x x 2x xy 1
x x 1 x
+ += + -
- +.
a) Rt gn y. Tm x y = 2. b) Gi s x > 1. Chng minh rng : y - | y | = 0 c) Tm gi tr nh nht ca y ?
PHN II: HNG DN GII
1. Gi s 7 l s hu t m7n
= (ti gin). Suy ra 2
2 22
m7 hay 7n m
n= = (1). ng thc
ny chng t 2m 7M m 7 l s nguyn t nn m M 7. t m = 7k (k Z), ta c m2 = 49k2 (2). T (1) v (2) suy ra 7n2 = 49k2 nn n2 = 7k2 (3). T (3) ta li c n2 M 7 v v 7 l s nguyn t nn n M 7. m v n cng chia ht cho 7 nn phn s m
n khng ti gin, tri gi thit. Vy 7
khng phi l s hu t; do 7 l s v t. 2. Khai trin v tri v t nhn t chung, ta c v phi. T a) b) v (ad bc)2 0. 3. Cch 1 : T x + y = 2 ta c y = 2 x. Do : S = x2 + (2 x)2 = 2(x 1)2 + 2 2. Vy min S = 2 x = y = 1.
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Cch 2 : p dng bt ng thc Bunhiacopxki vi a = x, c = 1, b = y, d = 1, ta c : (x + y)2 (x2 + y2)(1 + 1) 4 2(x2 + y2) = 2S S 2. mim S = 2 khi x = y = 1 4. b) p dng bt ng thc Cauchy cho cc cp s dng bc ca bc ab ca abv ; v ; v
a b a c b c,
ta ln lt c: bc ca bc ca bc ab bc ab
2 . 2c; 2 . 2ba b a b a c a c+ = + = ;
ca ab ca ab2 . 2a
b c b c+ = cng tng
v ta c bt ng thc cn chng minh. Du bng xy ra khi a = b = c. c) Vi cc s dng 3a v 5b , theo bt ng thc Cauchy ta c : 3a 5b 3a.5b
2+
.
(3a + 5b)2 4.15P (v P = a.b) 122 60P P 125
max P = 125
.
Du bng xy ra khi 3a = 5b = 12 : 2 a = 2 ; b = 6/5. 5. Ta c b = 1 a, do M = a3 + (1 a)3 = 3(a )2 + . Du = xy ra khi a = . Vy min M = a = b = . 6. t a = 1 + x b3 = 2 a3 = 2 (1 + x)3 = 1 3x 3x2 x3 1 3x + 3x2 x3 = (1 x)3. Suy ra : b 1 x. Ta li c a = 1 + x, nn : a + b 1 + x + 1 x = 2. Vi a = 1, b = 1 th a3 + b3 = 2 v a + b = 2. Vy max N = 2 khi a = b = 1. 7. Hiu ca v tri v v phi bng (a b)2(a + b). 8. V | a + b | 0 , | a b | 0 , nn : | a + b | > | a b | a2 + 2ab + b2 a2 2ab + b2 4ab > 0 ab > 0. Vy a v b l hai s cng du. 9. a) Xt hiu : (a + 1)2 4a = a2 + 2a + 1 4a = a2 2a + 1 = (a 1)2 0. b) Ta c : (a + 1)2 4a ; (b + 1)2 4b ; (c + 1)2 4c v cc bt ng thc ny c hai v u dng, nn : [(a + 1)(b + 1)(c + 1)]2 64abc = 64.1 = 82. Vy (a + 1)(b + 1)(c + 1) 8. 10. a) Ta c : (a + b)2 + (a b)2 = 2(a2 + b2). Do (a b)2 0, nn (a + b) 2 2(a2 + b2). b) Xt : (a + b + c)2 + (a b)2 + (a c)2 + (b c)2. Khai trin v rt gn, ta c : 3(a2 + b2 + c2). Vy : (a + b + c)2 3(a2 + b2 + c2).
11. a) 4
2x 3 1 x 3x 4 x2x 3 1 x 3
2x 3 x 1 x 2x 2
- = - = = - = - - = - = =
b) x2 4x 5 (x 2)2 33 | x 2 | 3 -3 x 2 3 -1 x 5. c) 2x(2x 1) 2x 1 (2x 1)2 0. Nhng (2x 1)2 0, nn ch c th : 2x 1 = 0 Vy : x = . 12. Vit ng thc cho di dng : a2 + b2 + c2 + d2 ab ac ad = 0 (1). Nhn hai v ca (1) vi 4 ri a v dng : a2 + (a 2b)2 + (a 2c)2 + (a 2d)2 = 0 (2). Do ta c :
a = a 2b = a 2c = a 2d = 0 . Suy ra : a = b = c = d = 0. 13. 2M = (a + b 2)2 + (a 1)2 + (b 1)2 + 2.1998 2.1998 M 1998.
Du = xy ra khi c ng thi : a b 2 0
a 1 0
b 1 0
+ - = - = - =
Vy min M = 1998 a = b = 1.
14. Gii tng t bi 13. 15. a ng thc cho v dng : (x 1)2 + 4(y 1)2 + (x 3)2 + 1 = 0. 16.
( )221 1 1 1
A . max A= x 2x 4x 9 5 5x 2 5
= = =- + - +
.
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17. a) 7 15 9 16 3 4 7+ < + = + = . Vy 7 15+ < 7 b) 17 5 1 16 4 1 4 2 1 7 49 45+ + > + + = + + = = > .
c) 23 2 19 23 2 16 23 2.4
5 25 273 3 3
- - -< = = = < .
d) Gi s
( ) ( )2 23 2 2 3 3 2 2 3 3 2 2 3 18 12 18 12> > > > > . Bt ng thc cui cng ng, nn : 3 2 2 3> .
18. Cc s c th l 1,42 v 2 32+
19. Vit li phng trnh di dng : 2 2 23(x 1) 4 5(x 1) 16 6 (x 1)+ + + + + = - + . V tri ca phng trnh khng nh hn 6, cn v phi khng ln hn 6. Vy ng thc ch xy ra khi c hai v u bng 6, suy ra x = -1.
20. Bt ng thc Cauchy a bab2+
vit li di dng 2
a bab
2+
(*) (a, b 0).
p dng bt dng thc Cauchy di dng (*) vi hai s dng 2x v xy ta c : 2
2x xy2x.xy 4
2+ =
Du = xy ra khi : 2x = xy = 4 : 2 tc l khi x = 1, y = 2. max A = 2 x = 2, y = 2. 21. Bt ng thc Cauchy vit li di dng : 1 2
a bab>
+. p dng ta c S > 19982.
1999.
22. Chng minh nh bi 1. 23. a)
2 2 2x y x y 2xy (x y)2 0
y x xy xy+ - -
+ - = = . Vy x y 2y x+
b) Ta c : 2 2 2 2
2 2 2 2
x y x y x y x y x yA 2
y x y x y x y x y x
= + - + = + - + + +
. Theo cu a :
2 22 2
2 2
x y x y x yA 2 2 1 1 0
y x y x y x
+ - + + = - + -
c) T cu b suy ra : 4 4 2 2
4 4 2 2
x y x y0
y x y x
+ - +
. V
x y2
y x+ (cu a). Do :
4 4 2 2
4 4 2 2
x y x y x y2
y x y x y x
+ - + + +
.
24. a) Gi s 1 2+ = m (m : s hu t) 2 = m2 1 2 l s hu t (v l)
b) Gi s m + 3n
= a (a : s hu t) 3n
= a m 3 = n(a m) 3 l s hu t, v l. 25. C, chng hn 2 (5 2) 5+ - =
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26. t 2 2
22 2
x y x ya 2 a
y x y x+ = + + = . D dng chng minh
2 2
2 2
x y2
y x+ nn a2 4, do
| a | 2 (1). Bt ng thc phi chng minh tng ng vi : a2 2 + 4 3a a2 3a + 2 0 (a 1)(a 2) 0 (2)
T (1) suy ra a 2 hoc a -2. Nu a 2 th (2) ng. Nu a -2 th (2) cng ng. Bi ton c chng minh. 27. Bt ng thc phi chng minh tng ng vi :
( )4 2 4 2 4 2 2 2 22 2 2
x z y x z x x z y x z y xyz0
x y z
+ + - + + .
Cn chng minh t khng m, tc l : x3z2(x y) + y3x2(y z) + z3y2(z x) 0. (1) Biu thc khng i khi hon v vng x y z x nn c th gi s x l s ln nht. Xt hai trng hp : a) x y z > 0. Tch z x (1) thnh (x y + y z), (1) tng ng vi :
x3z2(x y) + y3x2(y z) z3y2(x y) z3y2(y z) 0 z2(x y)(x3 y2z) + y2(y z)(yx2 z3) 0
D thy x y 0 , x3 y2z 0 , y z 0 , yx2 z3 0 nn bt ng thc trn ng. b) x z y > 0. Tch x y (1) thnh x z + z y , (1) tng ng vi :
x3z2(x z) + x3z2(z y) y3x2(z y) z3y2(x z) 0 z2(x z)(x3 zy2) + x2(xz2 y3)(z y) 0
D thy bt ng thc trn dng. Cch khc : Bin i bt ng thc phi chng minh tng ng vi :
2 2 2x y z x y z
1 1 1 3y z x y z x
- + - + - + + +
.
28. Chng minh bng phn chng. Gi s tng ca s hu t a vi s v t b l s hu t c. Ta c : b = c a. Ta thy, hiu ca hai s hu t c v a l s hu t, nn b l s hu t, tri vi gi thit. Vy c phi l s v t. 29. a) Ta c : (a + b)2 + (a b)2 = 2(a2 + b2) (a + b)2 2(a2 + b2). b) Xt : (a + b + c)2 + (a b)2 + (a c)2 + (b c)2. Khai trin v rt gn ta c : 3(a2 + b2 + c2). Vy : (a + b + c)2 3(a2 + b2 + c2) c) Tng t nh cu b 30. Gi s a + b > 2 (a + b)3 > 8 a3 + b3 + 3ab(a + b) > 8 2 + 3ab(a + b) > 8 ab(a + b) > 2 ab(a + b) > a3 + b3. Chia hai v cho s dng a + b : ab > a2 ab + b2 (a b)2 < 0, v l. Vy a + b 2. 31. Cch 1: Ta c : [ ]x x ; [ ]y y nn [ ]x + [ ]y x + y. Suy ra [ ]x + [ ]y l s nguyn khng vt qu x + y (1). Theo nh ngha phn nguyn, [ ]x y+ l s nguyn ln nht khng vt qu x + y (2). T (1) v (2) suy ra : [ ]x + [ ]y [ ]x y+ . Cch 2 : Theo nh ngha phn nguyn : 0 x - [ ]x < 1 ; 0 y - [ ]y < 1. Suy ra : 0 (x + y) ([ ]x + [ ]y ) < 2. Xt hai trng hp :
- Nu 0 (x + y) ([ ]x + [ ]y ) < 1 th [ ]x y+ = [ ]x + [ ]y (1) - Nu 1 (x + y) ([ ]x + [ ]y ) < 2 th 0 (x + y) ([ ]x + [ ]y + 1) < 1 nn
[ ]x y+ = [ ]x + [ ]y + 1 (2). Trong c hai trng hp ta u c : [ ]x + [ ]y [ ]x y+
-
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32. Ta c x2 6x + 17 = (x 3)2 + 8 8 nn t v mu ca A l cc s dng , suy ra A > 0 do : A ln nht 1
A nh nht x2 6x + 17 nh nht.
Vy max A = 18
x = 3.
33. Khng c dng php hon v vng quanh x y z x v gi s x y z. Cch 1 : p dng bt ng thc Cauchy cho 3 s dng x, y, z :
3x y z x y z
A 3 . . 3y z x y z x
= + + =
Do x y z x y zmin 3 x y zy z x y z x
+ + = = = = =
Cch 2 : Ta c : x y z x y y z yy z x y x z x x
+ + = + + + -
. Ta c x y 2y x+ (do x, y > 0) nn
chng minh x y z 3y z x+ + ta ch cn chng minh : y z y 1
z x x+ - (1)
(1) xy + z2 yz xz (nhn hai v vi s dng xz) xy + z2 yz xz 0 y(x z) z(x z) 0 (x z)(y z) 0 (2)
(2) ng vi gi thit rng z l s nh nht trong 3 s x, y, z, do (1) ng. T tm c gi tr nh nht ca x y z
y z x+ + .
34. Ta c x + y = 4 x2 + 2xy + y2 = 16. Ta li c (x y)2 0 x2 2xy + y2 0. T suy ra 2(x2 + y2) 16 x2 + y2 8. min A = 8 khi v ch khi x = y = 2. 35. p dng bt ng thc Cauchy cho ba s khng m :
1 = x + y + z 3. 3 xyz (1) 2 = (x + y) + (y + z) + (z + x) 3. 3 (x y)(y z)(z x)+ + + (2)
Nhn tng v ca (1) vi (2) (do hai v u khng m) : 2 9. 3 A A 3
29
max A = 3
29
khi v ch khi x = y = z = 13
.
36. a) C th. b, c) Khng th. 37. Hiu ca v tri v v phi bng (a b)2(a + b). 38. p dng bt ng thc 2
1 4xy (x y)
+
vi x, y > 0 : 2 2 2 2
2
a c a ad bc c 4(a ad bc c )b c d a (b c)(a d) (a b c d)
+ + + + + ++ =
+ + + + + + + (1)
Tng t 2 2
2
b d 4(b ab cd d )c d a b (a b c d)
+ + ++
+ + + + + (2)
Cng (1) vi (2) 2 2 2 2
2
a b c d 4(a b c d ad bc ab cd)b c c d d a a b (a b c d)
+ + + + + + ++ + +
+ + + + + + += 4B
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Cn chng minh B 12
, bt ng thc ny tng ng vi : 2B 1 2(a2 + b2 + c2 + d2 + ad + bc + ab + cd) (a + b + c + d)2 a2 + b2 + c2 + d2 2ac 2bd 0 (a c)2 + (b d)2 0 : ng.
39. - Nu 0 x - [ ]x < th 0 2x - 2[ ]x < 1 nn [ ]2x = 2[ ]x . - Nu x - [ ]x < 1 th 1 2x - 2[ ]x < 2 0 2x (2[ ]x + 1) < 1 [ ]2x = 2[ ]x + 1 40. Ta s chng minh tn ti cc s t nhin m, p sao cho :
14243mch so 0
96000...00 a + 15p < 14243mch so 0
97000...00
Tc l 96 +m ma 15p
10 10 < 97 (1). Gi a + 15 l s c k ch s : 10k 1 a + 15 < 10k
+ + + - + < + - . 49. A = 1 - | 1 3x | + | 3x 1 |2 = ( | 3x 1| - )2 + .
T suy ra : min A = x = hoc x = 1/6 51. M = 4
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52. x = 1 ; y = 2 ; z = -3.
53. P = | 5x 2 | + | 3 5x | | 5x 2 + 3 5x | = 1. min P = 1 2 3x5 5 .
54. Cn nh cch gii mt s phng trnh dng sau :
2
B 0A 0 (B 0) A 0a) A B b) A B c) A B 0
A B B 0A B
= = = + = = ==
B 0A 0
d) A B e) A B 0A BB 0
A B
== + = = = = -
.
a) a phng trnh v dng : A B= . b) a phng trnh v dng : A B= . c) Phng trnh c dng : A B 0+ = . d) a phng trnh v dng : A B= . e) a phng trnh v dng : | A | + | B | = 0 g, h, i) Phng trnh v nghim. k) t x 1- = y 0, a phng trnh v dng : | y 2 | + | y 3 | = 1 . Xt du v tri. l) t : 8x 1 u 0 ; 3x 5 v 0 ; 7x 4 z 0 ; 2x 2 t 0+ = - = + = - = .
Ta c h : 2 2 2 2
u v z t
u v z t
+ = +
- = -. T suy ra : u = z tc l : 8x 1 7x 4 x 3+ = + = .
55. Cch 1 : Xt 2 2 2 2 2x y 2 2(x y) x y 2 2(x y) 2 2xy (x y 2) 0+ - - = + - - + - = - - .
Cch 2 : Bin i tng ng ( )( )
22 22 2
2
x yx y2 2 8
x y x y
++
- - (x2 + y2)2 8(x y)2 0
(x2 + y2)2 8(x2 + y2 2) 0 (x2 + y2)2 8(x2 + y2) + 16 0 (x2 + y2 4)2 0. Cch 3 : S dng bt ng thc Cauchy :
2 2 2 2 2x y x y 2xy 2xy (x y) 2.1 2 1(x y) 2 (x y).
x y x y x y x y x y+ + - + - +
= = = - + -- - - - -
(x >
y).
Du ng thc xy ra khi 6 2 6 2x ; y2 2+ -
= = hoc 6 2 6 2x ; y2 2
- + - -= =
62. 2
2 2 2 2 2 2
1 1 1 1 1 1 1 1 1 1 1 1 2(c b a2a b c a b c ab bc ca a b c abc
+ + + + = + + + + + = + + +
=
= 2 2 21 1 1a b c+ + . Suy ra iu phi chng minh.
63. iu kin : 2 x 6(x 6)(x 10) 0x 16x 60 0
x 10x 10x 6x 6 0
x 6
- - - + -
.
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Bnh phng hai v : x2 16x + 60 < x2 12x + 36 x > 6. Nghim ca bt phng trnh cho : x 10. 64. iu kin x2 3. Chuyn v : 2x 3- x2 3 (1)
t tha chung : 2x 3- .(1 - 2x 3- ) 0 2
2
x 3x 3 0
x 21 x 3 0 x 2
= - =
- - -
Vy nghim ca bt phng trnh : x = 3 ; x 2 ; x -2. 65. Ta c x2(x2 + 2y2 3) + (y2 2)2 = 1 (x2 + y2)2 4(x2 + y2) + 3 = - x2 0. Do : A2 4A + 3 0 (A 1)(A 3) 0 1 A 3. min A = 1 x = 0, khi y = 1. max A = 3 x = 0, khi y = 3 . 66. a) x 1. b) B c ngha
2
2
2
4 x 44 x 416 x 0
x 4 2 2 12x 1 0 (x 4) 8 x 4 2 22x 4 2 21x 8x 8 0 x 12 x
2
- - - - + > - - < - + - + > -
> -
.
67. a) A c ngha 2
2 22
x 2x 0 x(x 2) 0 x 2x 0x x 2xx x 2x
- -
-
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T (1) , (2) : min A = 13
x = y = z = 33
71. Lm nh bi 8c ( 2). Thay v so snh n n 2 v 2 n+1+ + ta so snh n 2 n 1+ - + v n 1 n+ - . Ta c : n 2 n 1 n 1 n n n 2 2 n 1+ - + < + - + + < + .
72. Cch 1 : Vit cc biu thc di du cn thnh bnh phng ca mt tng hoc mt hiu. Cch 2 : Tnh A2 ri suy ra A. 73. p dng : (a + b)(a b) = a2 b2. 74. Ta chng minh bng phn chng. a) Gi s tn ti s hu t r m 3 5+ = r 3 + 2 15 + 5 = r2
2r 815
2-
= . V tri
l s v t, v phi l s hu t, v l. Vy 3 5+ l s v t. b), c) Gii tng t. 75. a) Gi s a > b ri bin i tng ng : 3 3 3 2 2 1 3 3 2 2 2= > - > + ( ) ( )2 23 3 2 2 2 27 8 4 8 2 15 8 2 225 128> + > + + > > . Vy a > b l ng. b) Bnh phng hai v ln ri so snh. 76. Cch 1 : t A = 4 7 4 7+ - - , r rng A > 0 v A2 = 2 A = 2 Cch 2 : t B = 4 7 4 7 2 2.B 8 2 7 8 2 7 2 0+ - - - = + - - - = B = 0.
77. ( ) ( )2 3 4 2 2 3 42 3 2.3 2.4 2 4
Q 1 22 3 4 2 3 4
+ + + + ++ + + += = = +
+ + + +.
78. Vit 40 2 2.5 ; 56 2 2.7 ; 140 2 5.7= = = . Vy P = 2 5 7+ + . 79. T gi thit ta c : 2 2x 1 y 1 y 1 x- = - - . Bnh phng hai v ca ng thc ny ta c : 2y 1 x= - . T : x2 + y2 = 1. 80. Xt A2 suy ra : 2 A2 4. Vy : min A = 2 x = 1 ; max A = 2 x = 0. 81. Ta c : ( ) ( ) ( )2 2 2M a b a b a b 2a 2b 2= + + + - = + .
1a bmax M 2 a b
2a b 1
== = =+ =
.
82. Xt tng ca hai s : ( ) ( ) ( ) ( )2a b 2 cd 2c d 2 ab a b 2 ab c d 2 cd a c+ - + + - = + - + + - + + = = ( ) ( ) ( )2 2a c a b c d a c 0+ + - + - + > . 83. N 4 6 8 3 4 2 18 12 8 3 4 4 6 4 2 2= + + + = + + + + + =
= ( ) ( ) ( )2 22 3 2 2 2 2 3 2 2 2 3 2 2 2 3 2 2+ + + + = + + = + + . 84. T x y z xy yz zx+ + = + + ( ) ( ) ( )2 2 2x y y z z x 0- + - + - = .
-
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Vy x = y = z. 85. p dng bt ng thc Cauchy cho 1 v ai ( i = 1, 2, 3, n ). 86. p dng bt ng thc Cauchy vi hai s a + b 0 v 2 ab 0, ta c :
( )2a b 2 ab 2 2(a b) ab hay a b 2 2(a b) ab+ + + + + . Du = xy ra khi a = b.
87. Gi s a b c > 0. Ta c b + c > a nn b + c + 2 bc > a hay ( ) ( )2 2b c a+ > Do : b c a+ > . Vy ba on thng a , b , c lp c thnh mt tam gic. 88. a) iu kin : ab 0 ; b 0. Xt hai trng hp : * Trng hp 1 : a 0 ; b > 0 : b.( a b) a a b aA 1
b bb. b b
- -= - = - = - .
* Trng hp 2 : a 0 ; b < 0 : 2
2
ab b a a a aA 1 1 2
b b b bb
-= - = - + - = -
-.
b) iu kin : 2(x 2) 8x 0
x 0x 0
x 22
x 0x
+ - > > -
. Vi cc iu kin th :
2 2 x 2 . x(x 2) 8x (x 2) . xB
2 x 2 x 2xx
-+ - -= = =
- --.
Nu 0 < x < 2 th | x 2 | = -(x 2) v B = - x . Nu x > 2 th | x 2 | = x 2 v B = x
89. Ta c : ( )222
2
2 2 2
a 1 1a 2 1a 1
a 1 a 1 a 1
+ ++= = + +
+ + +. p dng bt ng thc Cauchy:
2 2
2 2
1 1a 1 2 a 1. 2
a 1 a 1+ + + =
+ +. Vy
2
2
a 22
a 1
+
+. ng thc xy ra khi :
2
2
1a 1 a 0
a 1+ = =
+.
93. Nhn 2 v ca pt vi 2 , ta c : 2x 5 3 2x 5 1 4- + + - - = 5/2 x 3. 94. Ta chng minh bng qui np ton hc : a) Vi n = 1 ta c : 1
1 1P
2 3= < (*) ng.
b) Gi s : k1 1.3.5...(2k 1) 1
P2.4.6...2k2k 1 2k 1
-< + + - + +
x 03
x a340 x a
4
<
.
207. c) Trc ht tnh x theo a c 1 2ax2 a(1 a)
-=
-. Sau tnh 21 x+ c 1
2 a(1 a)-.
p s : B = 1. d) Ta c a2 + 1 = a2 + ab + bc + ca = (a + b)(a + c). Tng t :
b2 + 1 = (b + a)(b + c) ; c2 + 1 = (c + a)(c + b). p s : M = 0. 208. Gi v tri l A > 0. Ta c 2 2x 4A
x
+= . Suy ra iu phi chng minh.
209. Ta c : a + b = - 1 , ab = - 14
nn : a2 + b2 = (a + b)2 2ab = 1 + 1 32 2= .
a4 + b4 = (a2 + b2)2 2a2b2 = 9 1 174 9 8- = ; a3 + b3 = (a + b)3 3ab(a + b) = - 1 -
3 74 4= -
Do : a7 + b7 = (a3 + b3)(a4 + b4) a3b3(a + b) = ( )7 17 1 239. 14 8 64 64
- - - - = -
.
210. a) 2 2a ( 2 1) 3 2 2 9 8= - = - = - . 3 3a ( 2 1) 2 2 6 3 2 1 5 2 7 50 49= - = - + - = - = - .
b) Theo khai trin Newton : (1 - 2 )n = A - B 2 ; (1 + 2 )n = A + B 2 vi A, B N Suy ra : A2 2B2 = (A + B 2 )(A - B 2 ) = [(1 + 2 )(1 - 2 )]n = (- 1)n. Nu n chn th A2 2b2 = 1 (1). Nu n l th A2 2B2 = - 1 (2).
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WWW.MATHVN.COM MAI TRNG MU
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By gi ta xt an. C hai trng hp : * Nu n chn th : an = ( 2 - 1)n = (1 - 2 )n = A - B 2 = 2 2A 2B- . iu kin A2 2B2 = 1 c tha mn do (1). * Nu n l th : an = ( 2 - 1)n = - (1 - 2 )n = B 2 - A = 2 22B A- . iu kin 2B2 A2 = 1 c tha mn do (2). 211. Thay a = 2 vo phng trnh cho : 2 2 + 2a + b 2 + c = 0
2 (b + 2) = -(2a + c). Do a, b, c hu t nn phi c b + 2 = 0 do 2a + c = 0. Thay b = - 2 , c = - 2a vo phng trnh cho :
x3 + ax2 2x 2a = 0 x(x2 2) + a(x2 2) = 0 (x2 2)(x + a) = 0. Cc nghim phng trnh cho l: 2 v - a.
212. t 1 1 1A ...2 3 n
= + + + .
a) Chng minh A 2 n 3> - : Lm gim mi s hng ca A :
( )1 2 2 2 k 1 kk k k k 1 k= > = + -
+ + + .
Do ( ) ( ) ( )A 2 2 3 3 4 ... n n 1 > - + + - + + + - + + = ( )2 n 1 2 2 n 1 2 2 2 n 1 3 2 n 3= + - = + - > + - > - .
b) Chng minh A 2 n 2< - : Lm tri mi s hng ca A :
( )1 2 2 2 k k 1k k k k k 1= < = - -
+ + -
Do : ( ) ( ) ( )A 2 n n 1 ... 3 2 2 1 2 n 2 < - - + + - + - = - . 213. K hiu na 6 6 ... 6 6= + + + + c n du cn. Ta c :
1 2 1 3 2 100 99a 6 3 ; a 6 a 6 3 3 ; a 6 a 6 3 3 ... a 6 a 6 3 3= < = + < + = = + < + = = + < + =
Hin nhin a100 > 6 > 2. Nh vy 2 < a100 < 3, do [ a100 ] = 2. 214. a) Cch 1 (tnh trc tip) : a2 = (2 + 3 )2 = 7 + 4 3 . Ta c 4 3 48= nn 6 < 4 3 < 7 13 < a2 < 14. Vy [ a2 ] = 13. Cch 2 (tnh gin tip) : t x = (2 + 3 )2 th x = 7 + 4 3 . Xt biu thc y = (2 - 3 )2 th y = 7 - 4 3 . Suy ra x + y = 14. D thy 0 < 2 - 3 < 1 nn 0 < (2- 3 )2 < 1, tc l 0 < y < 1. Do 13 < x < 14.
Vy [ x ] = 13 tc l [ a2 ] = 13. b) p s : [ a3 ] = 51. 215. t x y = a ; x y b+ = (1) th a v b l s hu t. Xt hai trng hp :
a) Nu b 0 th x y a ax yb bx y
-= - =
+ l s hu t (2). T (1) v (2) ta c :
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WWW.MATHVN.COM MAI TRNG MU
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1 ax b
2 b = +
l s hu t ; 1 ay b2 b = -
l s hu t.
b) Nu b = 0 th x = y = 0, hin nhin x , y l s hu t.
216. Ta c 1 n 1 1 1 1 1 1
n nn(n 1) n n 1(n 1) n n n 1 n n 1
= = - = + - = + ++ + +
n 1 1 1 11 2
n 1 n n 1 n n 1
= + - < - + + + . T ta gii c bi ton.
217. Chng minh bng phn chng. Gi s trong 25 s t nhin cho, khng c hai s no bng nhau. Khng mt tnh tng qut, gi s a1 < a2 < . < a25. Suy ra : a1 1 , a2 2 , a25 25. Th th :
1 2 25
1 1 1 1 1 1.... ....
a a a 1 2 25+ + + + + + (1). Ta li c :
1 1 1 1 2 2 2.... .... 1
25 24 2 1 25 25 24 24 2 2+ + + + = + + + + 0 T h phng trnh cho ta c : 2y 2yx y
1 y 2 y= =
+.
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Tng t y z ; z x . Suy ra x = y = z. Xy ra du = cc bt ng thc trn vi x = y = z = 1. Kt lun : Hai nghim (0 ; 0 ; 0) , (1 ; 1 ; 1). 221. a) t A = (8 + 3 7 )7. chng minh bi ton, ch cn tm s B sao cho 0 < B < 7
110
v A + B l s t nhin. Chn B = (8 - 3 7 )7. D thy B > 0 v 8 > 3 7 . Ta c 8 + 3 7 > 10 suy ra :
( )( )77 7 71 1 18 3 710 108 3 7
< - 3, khi A 0 (2). So snh (1) v (2) ta i n kt lun :
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x 3 xmaxA 4 x 22
x 0
= -= =
.
229. a) Lp phng hai v, p dng hng ng thc (a + b)3 = a3 + b3 + 3ab(a + b), ta c : 3x 1 7 x 3. (x 1)(7 x).2 8 (x 1)(7 x) 0+ + - + + - = + - = x = - 1 ; x = 7 (tha)
b) iu kin : x - 1 (1). t 3 x 2 y ; x 1 z- = + = . Khi x 2 = y2 ; x + 1 = z2 nn z2 y3 = 3. Phng trnh cho c a v h :
2 3
y z 3 (2)
z y 3 (3)z 0 (4)
+ = - =
Rt z t (2) : z = 3 y. Thay vo (3) : y3 y2 + 6y 6 = 0 (y 1)(y2 + 6) = 0 y = 1 Suy ra z = 2, tha mn (4). T x = 3, tha mn (1). Kt lun : x = 3. 230. a) C, chng hn : 1 1 2
2 2+ = .
b) Khng. Gi s tn ti cc s hu t dng a, b m 4a b 2+ = . Bnh phng hai v : a b 2 ab 2 2 ab 2 (a b)+ + = = - + .
Bnh phng 2 v : 4ab = 2 + (a + b)2 2(a + b) 2 2(a + b) 2 = 2 + (a + b)2 4ab V phi l s hu t, v tri l s v t (v a + b 0), mu thun. 231. a) Gi s 3 5 l s hu t m
n (phn s ti gin). Suy ra 5 =
3
3
mn
. Hy chng minh rng
c m ln n u chia ht cho 5, tri gi thit mn
l phn s ti gin.
b) Gi s 3 32 4+ l s hu t mn
(phn s ti gin). Suy ra :
( )3 3 3 3 2 33 3 33
m m 6m2 4 6 3. 8. 6 m 6n 6mn (1) m 2 m 2n n n
= + = + = + = + M M
Thay m = 2k (k Z) vo (1) : 8k3 = 6n3 + 12kn2 4k3 = 3n3 + 6kn2. Suy ra 3n3 chia ht cho 2 n3 chia ht cho 2 n chia ht cho 2. Nh vy m v n cng chia ht cho 2, tri vi gi thit m
n l phn s ti gin.
232. Cch 1 : t a = x3 , b = y3 , c = z3. Bt ng thc cn chng minh 3a b c abc3
+ +
tng ng vi 3 3 3x y z xyz hay
3+ +
x3 + y3 + z3 3xyz 0. Ta c hng ng thc :
x3 + y3 + z3 3xyz = 12
(x + y + z)[(x y)2 + (y z)2 + (z x)2]. (bi tp sbt)
Do a, b, c 0 nn x, y, z 0, do x3 + y3 + z3 3xyz 0. Nh vy : 3a b c abc3
+ +
Xy ra du ng thc khi v ch khi a = b = c. Cch 2 : Trc ht ta chng minh bt ng thc Cauchy cho bn s khng m. Ta c :
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( ) 4a b c d 1 a b c d 1 ab cd ab. cd abcd4 2 2 2 2+ + + + + = + + =
Trong bt ng thc 4a b c d abcd
4+ + +
, t a b cd
3+ +
= ta c : 4
4a b ca b c a b c a b c a b c3 abc. abc.
4 3 3 3
+ + + + + + + + + + +
.
Chia hai v cho s dng a b c3
+ + (trng hp mt trong cc s a, b, c bng 0, bi ton c
chng minh) : 3
3a b c a b cabc abc3 3
+ + + +
.
Xy ra ng thc : a = b = c = a b c3
+ + a = b = c = 1
233. T gi thit suy ra : b c d a 11b 1 c 1 d 1 a 1 a 1
+ + - =+ + + + +
. p dng bt ng thc
Cauchy cho 3 s dng : 31 b c d bcd3.a 1 b 1 c 1 d 1 (b 1)(c 1)(d 1)
+ + + + + + + + +
. Tng t :
3
3
3
1 acd3.b 1 (a 1)(c 1)(d 1)
1 abd3.c 1 (a 1)(b 1)(d 1)
1 abc3.d 1 (a 1)(b 1)(c 1)
+ + + +
+ + + +
+ + + +
Nhn t bn bt ng thc : 11 81abcd abcd81
.
234. Gi 2 2 2
2 2 2
x y zAy z x
= + + . p dng bt ng thc Bunhiacpxki : 22 2 2
2 2 2
x y z x y z3A (1 1 1)y z x y z x
= + + + + + +
(1)
p dng bt ng thc Cauchy vi ba s khng m : 3x y z x y z3. . . 3y z x y z x+ + = (2)
Nhn tng v (1) vi (2) : 2
x y z x y z x y z3A 3 Ay z x y z x y z x
+ + + + + +
235. t 3 33 3x 3 3 ; y 3 3= + = - th x3 + y3 = 6 (1). Xt hiu b3 a3 , ta c : b3 a3 = 24 (x + y)3 = 24 (x3 + y3) 3xy(x + y)
Do (1), ta thay 24 bi 4(x3 + b3), ta c :
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b3 a3 = 4(x3 + y3) (x3 + y3) 3xy(x + y) = 3(x3 + y3) 3xy(x + y) = = 3(x + y)(x2 xy + y2 xy) = 3(x + y)(x y)2 > 0 (v x > y > 0).
Vy b3 > a3 , do b > a. 236. a) Bt ng thc ng vi n = 1. Vi n 2, theo khai trin Newton, ta c :
n
2 3 n
1 1 n(n 1) 1 n(n 1)(n 2) 1 n(n 1)...2.1 11 1 n. . . ... .n n 2! n 3! n n! n
- - - - + = + + + + +
< 1 1 11 1 ...2! 3! n! + + + + +
D dng chng minh : 1 1 1 1 1 1... ...2! 3! n! 1.2 2.3 (n 1)n+ + + + + + =
-
= 1 1 1 1 1 11 ... 1 12 2 3 n 1 n n
- + - + + - = - (1). Tht vy, (1) ( ) ( )6 63 3 2> 32 > 22. Vi n 3, ta chng minh n n 1n n 1+> + (2). Tht vy :
( ) ( )nnn(n 1) n(n 1) n n 1n 1 n
n
(n 1) 1(2) n 1 n (n 1) n n 1 nn n
+ +++ + + < + < < +
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Cch 1 : Vi 0 x < 6 th A = x(x2 6) 0. Vi x 6 . Ta c 6 x 3 6 x2 9 0 x2 6 3. Suy ra x(x2 6) 9. max A = 9 vi x = 3. Cch 2 : A = x(x2 9) + 3x. Ta c x 0, x2 9 0, 3x 9, nn A 9.
max A = 9 vi x = 3 b) Tm gi tr nh nht : Cch 1 : A = x3 6x = x3 + (2 2 )3 6x (2 2 )3 =
= (x + 2 2 )(x2 - 2 2 x + 8) 6x - 16 2 = (x + 2 2 )(x2 - 2 2 x + 2) + (x + 2 2 ).6 6x - 16 2
= (x + 2 2 )(x - 2 )2 - 4 2 - 4 2 . min A = - 4 2 vi x = 2 .
Cch 2 : p dng bt ng thc Cauchy vi 3 s khng m : x3 + 2 2 + 2 2 3. 33 x .2 2.2 2 = 6x.
Suy ra x3 6x - 4 2 . min A = - 4 2 vi x = 2 . 241. Gi x l cnh ca hnh vung nh, V l th tch ca hnh hp. Cn tm gi tr ln nht ca V = x(3 2x)2. Theo bt ng thc Cauchy vi ba s dng :
4V = 4x(3 2x)(3 2x) 34x 3 2x 3 2x
3+ - + -
= 8
max V = 2 4x = 3 2x x = 12
Th tch ln nht ca hnh hp l 2 dm3 khi cnh hnh vung nh bng 12
dm.
242. a) p s : 24 ; - 11. b) t 3 2 x a ; x 1 b- = - = . p s : 1 ; 2 ; 10.
c) Lp phng hai v. p s : 0 ; 52
d) t 3 2x 1- = y. Gii h : x3 + 1 = 2y , y3 + 1 = 2x, c (x y)(x2 + xy + y2 + 2) = 0
x = y. p s : 1 ; 1 52
- .
e) Rt gn v tri c : ( )21 x x 42 - - . p s : x = 4. g) t 3 37 x a ; x 5 b- = - = . Ta c : a3 + b3 = 2, a3 b3 = 12 2x, do v phi ca
phng trnh cho l 3 3a b2- . Phng trnh cho tr thnh : a b
a b-+
= 3 3a b2-
.
Do a3 + b3 = 2 nn 3 3
3 3
a b a ba b a b- -
=+ +
(a b)(a3 + b3) = (a + b)(a3 b3)
Do a + b 0 nn : (a b)(a2 ab + b2 = (a b)(a2 + ab + b2). T a = b ta c x = 6. T ab = 0 ta c x = 7 ; x = 5. h) t 3 3x 1 a ; x 1 b+ = - = . Ta c : a2 + b2 + ab = 1 (1) ; a3 b3 = 2 (2). T (1) v (2) : a b = 2. Thay b = a 2 vo (1) ta c a = 1. p s : x = 0.
3-2x
3-2x
xx x
x
x
xx
x
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i) Cch 1 : x = - 2 nghim ng phng trnh. Vi x + 2 0, chia hai v cho 3 x 2+ .
t 3 x 1 x 3a ; bx 2 x 2+ +
= =+ +
. Gii h a3 + b3 = 2, a + b = - 1. H ny v nghim.
Cch 2 : t 3 x 2+ = y. Chuyn v : 3 33 3y 1 y 1 y- + + = - . Lp phng hai v ta c : y3 1 + y3 + 1 + 3. 63 y 1- .(- y) = - y3 y3 = y. 63 y 1- .
Vi y = 0, c nghim x = - 2. Vi y 0, c y2 = 63 y 1- . Lp phng : y6 = y6 1. V n0. Cch 3 : Ta thy x = - 2 nghim ng phng trnh. Vi x < - 2, x > - 2, phng trnh v nghim, xem bng di y :
x 3 x 1+ 3 x 2+ 3 x 3+ V tri x < - 2 x > - x
< - 1 > - 1
< 0 > 0
< 1 > 1
< 0 > 0
k) t 1 + x = a , 1 x = b. Ta c : a + b = 2 (1), 4 4 4ab a b+ + = 3 (2) Theo bt ng thc Cauchy m nmn
2+
, ta c :
a b 1 a 1 b3 a. b 1. a 1. b2 2 2+ + +
= + + + + =
1 a 1 b a ba b 1 1 2 32 2 2+ + +
= + + + + = + = .
Phi xy ra du ng thc, tc l : a = b = 1. Do x = 0. l) t 4 4a x m 0 ; b x n 0- = - = th m4 + n4 = a + b 2x. Phng trnh cho tr thnh : m + n = 4 44 m n+ . Nng ln ly tha bc bn hai v ri thu gn : 2mn(2m2 + 3mn + 2n2) = 0. Suy ra m = 0 hoc n = 0, cn nu m, n > 0 th 2m2 + 3mn + 2n2 > 0. Do x = a , x = b. Ta phi c x a , x b cc cn thc c ngha. Gi s a b th nghim ca phng trnh cho l x = a. 243. iu kin biu thc c ngha : a2 + b2 0 (a v b khng ng thi bng 0). t 3 3a x ; b y= = , ta c :
4 2 2 4 4 2 2 4 2 2
2 2 2 2
x x y y x 2x y y 2x yA
x xy y x xy y+ + + + -
= =+ + + +
=
( ) ( )( )22 2 2 2 2 2 2 2 22 2 2 2
x y (xy) x y xy x y xyx y xy
x xy y x y xy
+ - + + + -= = = + -
+ + + +.
Vy : 2 23 3 3A a b ab= + - (vi a2 + b2 0). 244. Do A l tng ca hai biu thc dng nn ta c th p dng bt ng thc Cauchy :
2 2 2 2 2 24A x x 1 x x 1 2 x x 1. x x 1 2 (x x 1)(x x 1)= - + + + + - + + + = - + + + =
= 4 242 x x 2 2+ + . ng thc xy ra khi : 2 2
4 2
x x 1 x x 1x 0
x x 1 1
+ + = - + =+ + =
.
Ta c A 2, ng thc xy ra khi x = 0. Vy : min A = 2 x = 0.
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245. V 1 + 3 l nghim ca phng trnh 3x3 + ax2 + bx + 12 = 0, nn ta c : 3(1 + 3 )3 + a(1 + 3 )2 + b(1 + 3 ) + 12 = 0.
Sau khi thc hin cc php bin i, ta c biu thc thu gn : (4a + b + 42) + (2a + b + 18) 3 = 0.
V a, b Z nn p = 4a + b + 42 Z v q = 2a + b + 18 Z. Ta phi tm cc s nguyn a, b sao cho p + q 3 = 0.
Nu q 0 th 3 = - pq
, v l. Do q = 0 v t p + q 3 = 0 ta suy ra p = 0.
Vy 1 + 3 l mt nghim ca phng trnh 3x3 + ax2 + bx + 12 = 0 khi v ch khi : 4a b 42 0
2a b 18 0
+ + = + + =
. Suy ra a = - 12 ; b = 6.
246. Gi s 3 3 l s hu t pq
(pq
l phn s ti gin ). Suy ra : 3 = 3
3
pq
. Hy chng minh c p
v q cng chia ht cho 3, tri vi gi thit pq
l phn s ti gin.
247. a) Ta c : ( )23 6 661 2 1 2 1 2 2 2 3 2 2+ = + = + + = + . Do : ( )223 6 6 6 61 2 . 3 2 2 3 2 2. 3 2 2 3 2 2 1+ - = + - = - = . b) 6 39 4 5. 2 5 1+ - = - . 248. p dng hng ng thc (a + b)3 = a3 + b3 + 3ab(a + b), ta c :
3 3 2 23 3a 20 14 2 20 14 2 3 (20 14 2)(20 14 2).a a 40 3 20 (14 2) .a= + + - + + - = + - a3 6a 40 = 0 (a 4)(a2 + 4a + 10) = 0. V a2 + 4a + 10 > 0 nn a = 4. 249. Gii tng t bi 21. 250. A = 2 + 3 2- . 251. p dng : (a + b)3 = a3 + b3 + 3ab(a + b).
T x = 3 33 9+ . Suy ra x3 = 12 + 3.3x x3 9x 12 = 0. 252. S dng hng ng thc (A B)3 = A3 B3 3AB(A B). Tnh x3. Kt qu M = 0 253. a) x1 = - 2 ; x2 = 25.
b) t 3u x 9 , v x 3= - = - , ta c : 3
3
u v 6
v u 6
= +
= + u = v = - 2 x = 1.
c) t : 24 x 32 y 0+ = > . Kt qu x = 7. 254. a biu thc v dng : 3 3A x 1 1 x 1 1= + + + + - . p dng | A | + | B | | A + B |
min A = 2 -1 x 0. 255. p dng bt ng thc Cauchy hai ln. 256. t 3 2 23 3x y th x y P 2 x 2= = = + 258. Ta c : ( ) ( )2 2P x a x b= - + - = | x a | + | x b | | x a + b x | = b a (a < b).
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Du ng thc xy ra khi (x a)(x b) 0 a x b. Vy min P = b a a x b. 259. V a + b > c ; b + c > a ; c + a > b. p dng bt ng thc Cauchy cho tng cp s dng
(a b c) (b c a)(a b c)(b c a) b
2(b c a) (c a b)
(b c a)(c a b) c2
(c a b) (a b c)(c a b)(a b c) a
2
+ - + + -+ - + - =
+ - + + -+ - + - =
+ - + + -+ - + - =
Cc v ca 3 bt dng thc trn u dng. Nhn 3 bt ng thc ny theo tng v ta c bt ng thc cn chng minh. ng thc xy ra khi v ch khi :
a + b c = b + c a = c + a b a = b = c (tam gic u). 260. 2 2x y (x y) (x y) 4xy 4 4 2 2- = - = + - = + = . 261. 2A = (a b)2 + (b c)2 + (c a)2.
Ta c : c a = - (a c) = - [(a b) + (b c)] = - ( 2 + 1 + 2 - 1) = - 2 2 .
Do : 2A = ( 2 + 1)2 + ( 2 - 1)2 + (-2 2 )2 = 14. Suy ra A = 7. 262. a pt v dng : ( ) ( ) ( )2 2 2x 2 1 y 3 2 z 5 3 0- - + - - + - - = . 263. Nu 1 x 2 th y = 2. 264. t : ( )( )x 1 y 0. M x 1 x 1 2 3 x 1- = = - - + - - . 265. Gi cc kch thc ca hnh ch nht l x, y. Vi mi x, y ta c : x2 + y2 2xy. Nhng x2 + y2 = (8 2 )2 = 128, nn xy 64. Do : max xy = 64 x = y = 8. 266. Vi mi a, b ta lun c : a2 + b2 2ab. Nhng a2 + b2 = c2 (nh l Pytago) nn : c2 2ab 2c2 a2 +b2 + 2ab 2c2 (a + b)2 c 2 a + b c a b
2
+.
Du ng thc xy ra khi v ch khi a = b. 267. Bin i ta c : ( ) ( ) ( )2 2 2a 'b ab ' a 'c ac ' b 'c bc ' 0- + - + - = 268. 2 x - 1 ; 1 x 2.
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