Phương trrình, hệ phương trình

BAI TAP HE PHONG TRINH

Giải các hệ phơng trình sau :

‘i +xy+y=-l

(MTCN —99) x2y+y x=-6

x+y =5

2, x -=xy +y =l3 4 2.2 4 (NT — 98)

—+,/—- =——+1

| Xa xy + yrafxy = 78

1 1

x+y+—+—=4

x+y? tatty

x +x4+

Jer txtytltxtaly? +x+y+Il+y=18 y y y y (AN -99)

x

Jetxtytl— x+Ay' +x+y+l-y=2

12,40 PCED" ecvr -97) 13,47 `” (SP1—2000)

(x? ty" )O" ty") = 280 2y -3y=x”-2

2x+l =3

xy

3

2x+y=—

y

22,4, (HH —TPHCM) (TM — 2001)

x” —3xy—-2x7 =8

2, » 23, y

ytxy? =-6x7

2x° —13xy+15y? =0 x(x° + y7)=10y

xy+x+l=7y

26,

xy +xy+l=13y

x(x+y+l1)—-3=0

(ĐH Khối D -2009) DS;

(x+y) -2.415 0

xVx+y.Jy =l—3m 4

—

27, Tim m đê hệ sau có nghiệm: Ị Ị ĐS:— <”m <2,m 3 22

x`+-c+yÌ+—=157m—10

28 Giải ph ư ơng tr ình sau: 2Ä3x—2+3/6—5x—8=0 DS: x=-2

BAI TAP PHONG TRINH -BAT PHONG TRINH VO TI

Giải các phơng trình sau:

7.2A|x+2+Ax+l-jx+1=4(@hối— 2005) — 8,

\\x+2Ax—1 —A|x—2AÍx—1 =2(BCV7 —2000)

9,3(2+Ax-2)=2x+Ax+6(HVKT0S~01)

10,42x?+8x+6+Ax?—1=2x+2(8K —2000)

Joe evi poe vine = x + 1(PCCC —2001)

12, fx(x—D + fxr +2) = 2vx? (SP2—2000A)

13,J2x? +8x+6 +x? —1 =2x +2(HVKTOS —99)

Tim m dé phong trinh :

Giải các phơng trình sau :

17,x”+Ax °+11=3l1

19,4x?—3x+3+Ala?—3x+6 =3(TM —98)

21,47 +2x4+4=3Vx° +45

\3-x+a? -J2+x-x? = 1(NT -—99)

18,(x+5)(2—x) =3Vx° +3x 20,2x° +5x-1=7Vx-1

22,

23,Vx+1+V4—x + (44+ D(4—x)(NN — 20001)

24,x+V4—x? =24+3xV4—2? (MDC —2001)

25,Vx-24+V4-x =x? -6x411

26, J2x-34+V5—-2x +4x—x? —6 = 0(GTVT -TPHCM - 01)

27,V3x—-24Vx-1=4x-942V3x° —5x+2(HVKTOS —97)

“+ 7x+4

28, ——— =4Ax(DL Đông Đô-2000)

X+

30,* + = =242

x-Il

32,(4x—I)Nx?+1=2x?+2x+1@п78)

34.2(1—x)Nxˆ+2x-l=x”-2x-—I

36,82 x =1—AJx—I(TCKT - 2000)

Ñ7—x—ÄÍx—5

7—x+Äx—5

Giải các bất phơng trình sau :

1L.(x—(4—x) >x—2(MĐC-—2000)

3,Vx+3 >V2x-8 +V7—x(AN —-97)

5,(x-3)W.x? —4 <x? -9(Dé11)

38, =6-—x(C D- KiémSat)

> x—4(§PVinh —01)

2

XxX

1,

d+Vx+1)

29, = +3 a4 = 2(GT —95)

31,j1+A41-x? =x+2N1-a?)

33,x“+3x+1=(x+3)N\x” +1(GT —01) 35,x° +Vx+1 =1(XD—98)

37,4Íx+7 —xÍx = 1(Luật -96)

39,x°+1=23/2x—I

2,Vx+1>3-Vx+4(BK —99)

4, x +2 —V3—x <V5—2x(TL—2000)

1-V1-4x

XxX g.12+x— x” > V124+x—-2x2

6 <3(NN —98)

(Hué —99)

9,4x?+3x+2 +43? +6x+5 S423? +9x+7(BK — 2000)

10, Vx? 4x43 —V2x? —3x +1 > x—1(KT —2001)

11,J5x7 +10x+12>7-—x° —2x(Dé135)

12, -44(4- x(2+x)<x?~2x—12(У149)

13,(0° +1) +(x? +1) 43xV x41 > 0(XD —99)