Bai Tap He Pt; Pt; Bpt

  • April 2020
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Bµi tËp hÖ ph¬ng tr×nh Gi¶i c¸c hÖ ph¬ng tr×nh sau :  x + xy + y = −1 (MTCN − 99) 1,  2 2 x y + y x = − 6   x2y + y2x = 30 (BK − 93) 3,  3 3  x + y = 35  x2 + y2 + xy = 7 (SP1− 2000) 5,  4 4 2 2  x + y + x y = 21

 x2 + y2 = 5 (NT − 98) 2,  4 2 2 4  x − x y + y = 13  x3 + y3 = 1 (AN − 97) 4,  5 5 2 2  x + y = x + y  x + y + xy = 11 (QG − 2000) 6,  2 2 x + y + 3( x + y ) = 28 

 x y 7 + = +1  x xy (HH − 99) 7,  y   x xy + y xy = 78

1  (x + y)(1+ xy) = 5  (NT − 99) 8,  (x2 + y2 )(1+ 1 ) = 49  x2y2

1 1   x + y+ x + y = 4  (AN − 99) 9,  1 1 2 2 x + y + + =4  x2 y2

 x(x + 2)(2x + y) = 9 (AN − 2001) 10,  2  x + 4x + y = 6

 x2 + x + y + 1 + x + y2 + x + y + 1 + y = 18 (AN − 99) 11,   x2 + x + y + 1 − x + y2 + x + y + 1 − y = 2  y + xy2 = 6x2  x(3x + 2y)(x + 1) = 12 ( BCVT − 97) (SP1− 2000) 12,  2 13,  2 2 2 x + 2 y + 4 x − 8 = 0 1 + x y = 5 x    2x2 − 3x = y2 − 2 x+ y = 4 ( HVQHQT − 2001) (QG − 2000) 14,  2 2 3 3 15,  2 2 ( x + y )( x + y ) = 280 2 y − 3 y = x − 2    1 3  2x + = 2  y x  x = 3x − y  (MTCN − 98) (QG − 99) 16,  2 17,  1 3  y = 3y − x 2y + =  x y  x = 3x + 8y (QG − 98) 18,  3  y = 3y + 8x 3

 x + 5 + y − 2 = 7 (NN1− 2000) 20,   y + 5 + x − 2 = 7

3  2x + y = x2 (TL − 2001) 19,  3 2y + x =  y2  y2 + 2 3 y =  x2  (KhèiB − 2003) 21,  2 3x = x + 2  y2

GV:NGUYỄN MINH NHIÊN-TRƯỜNG THPT QUẾ VÕ 1

3x2 − 2xy = 16 (HH − TPHCM ) 22,  2 2  x − 3xy − 2x = 8  x2 − 2xy + 3y2 = 9 (HVNH − TPHCM ) 24,  2 2 2x − 13xy + 15y = 0

1+ x3y3 = 19x3 (TM − 2001) 23,  2 2  y + xy = −6x 2y(x2 − y2 ) = 3x (M § C − 97) 25,  2 2  x(x + y ) = 10y

Bµi tËp ph¬ng tr×nh -bÊt ph¬ng tr×nh v« tØ Gi¶i c¸c ph¬ng tr×nh sau: 1, x + 3 + 6 − x = 3

2, x + 9 = 5− 2x + 4

3, x + 4 − 1− x = 1− 2x

4, (x − 3) 10 − x2 = x2 − x − 12

5, 3 x + 4 − 3 x − 3 = 1

6, 3 2x − 1 + 3 x − 1 = 3 3x + 1

7, 2 x + 2 + x + 1 − x + 1 = 4(khèiD − 2005)

8,

x + 2 x − 1 − x − 2 x − 1 = 2(BCVT − 2000) 9, 3(2 + x − 2) = 2x + x + 6(HVKTQS − 01) 10, 2x2 + 8x + 6 + x2 − 1 = 2x + 2(BK − 2000) 11,

5 2 5 2 − x + 1− x2 + − x − 1− x2 = x + 1(PCCC − 2001) 4 4

12, x(x − 1) + x(x + 2) = 2 x2 (SP 2 − 2000A) 13, 2x2 + 8x + 6 + x2 − 1 = 2x + 2(HVKTQS − 99) T×m m ®Ó ph¬ng tr×nh : 14, x2 + mx + 2 = 2x + 1(KhèiB − 2006) cã 2 nghiÖm ph©n biÖt 15, 2x2 + mx = 3− x(SPKT − TPHCM )

cã nghiÖm

16, 2x2 + mx − 3 = x − m(GT − 98) Gi¶i c¸c ph¬ng tr×nh sau : 17, x2 + x2 + 11 = 31

cã nghiÖm

19, x2 − 3x + 3 + x2 − 3x + 6 = 3(TM − 98) 21, x2 + 2x + 4 = 3 x3 + 4x

18, (x + 5)(2 − x) = 3 x2 + 3x 20, 2x2 + 5x − 1 = 7 x3 − 1 22,

3− x + x2 − 2+ x − x2 = 1(NT − 99) 23, x + 1 + 4 − x + (x + 1)(4 − x)(NN − 20001) 24, x + 4 − x2 = 2 + 3x 4− x2 (M § C − 2001) 25, x − 2 + 4 − x = x2 − 6x + 11 26, 2x − 3 + 5 − 2x + 4x − x2 − 6 = 0(GTVT − TPHCM − 01) 27, 3x − 2 + x − 1 = 4x − 9 + 2 3x2 − 5x + 2(HVKTQS − 97)

GV:NGUYỄN MINH NHIÊN-TRƯỜNG THPT QUẾ VÕ 1

x2 + 7x + 4 = 4 x(DL § «ng§ «−2000) x+ 2 2x 3 1 1 3 + + = 2(GT − 95) x+ 1 2 2x x =2 2 30, x + x2 − 1 28,

29,

31,

1+ 1− x2 = x(1+ 2 1− x2 ) 32, (4x − 1) x2 + 1 = 2x2 + 2x + 1(§ Ò78)

33,

x2 + 3x + 1 = (x + 3) x2 + 1(GT − 01) 34, 2(1− x) x2 + 2x − 1 = x2 − 2x − 1

35, x2 + x + 1 = 1( XD − 98)

36, 3 2 − x = 1− x − 1(TCKT − 2000) 38,

3

7− x − 3 x − 5

= 6− x(C § − KiÓmS¸t)

7− x + x − 5 Gi¶i c¸c bÊt ph¬ng tr×nh sau : 1, (x − 1)(4 − x) > x − 2(M § C − 2000) 3

3

3, x + 3 ≥ 2x − 8 + 7− x(AN − 97)

37, 3 x + 7 − x = 1(LuËt − 96) 39, x3 + 1 = 23 2x − 1 2, x + 1 > 3− x + 4(BK − 99) 4,

x + 2 − 3− x < 5− 2x(TL − 2000) 5, (x − 3) x2 − 4 ≤ x2 − 9(§ Ò11) 7,

x2 (1+ x + 1)2

6,

1− 1− 4x2 < 3(NN − 98) x

> x − 4(SPVinh − 01)

8,

12 + x − x2 12+ x − x2 ≥ (HuÕ − 99) x − 11 2x − 9 9, x2 + 3x + 2 + x2 + 6x + 5 ≤ 2x2 + 9x + 7(BK − 2000) 10, x2 − 4x + 3 − 2x2 − 3x + 1 ≥ x − 1(KT − 2001) 11, 5x2 + 10x + 1 ≥ 7− x2 − 2x(§ Ò135) 12, −4 (4 − x)(2 + x) ≤ x2 − 2x − 12(§ Ò149) 13, (x3 + 1) + (x2 + 1) + 3x x + 1 > 0( XD − 99) 3 1 < 2x + − 7(Th¸iNguyª n − 2000) 14, 3 x + 2x 2 x 15, x(x − 4) − x2 + 4x + (x − 2)2 < 2(HVNH − 99)

GV:NGUYỄN MINH NHIÊN-TRƯỜNG THPT QUẾ VÕ 1

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