TRUGNG THCS PHU THO QUAN 11 GV: CAO MINH TAI TRANG 1
CAC BAL TAP MAU
1/ x(x-1)(2-3x) =0
©x=0O0hay x- l =0 hay 2-3x =0
© x= 0 hay x = l hay —-3x =—2
© x = (hay x =| hay x= 5
1
2/ (2x +1)(9x?—25) =0
& (Sx +1)(3x—5)(3x+5) =0
1
© zx=-] hay 3x = 5 hay 3x =—5
2h _— h _ TS
©x=-2hayx= -hayX= 4
x 2
3) x(5-2)(x? 43) = 0
©x=0hay s~2=0hay x? +3 =0
© x= 0 hay 2= 2 hay x?=—3 (vô l0
©x=0Ohayx=Ó
4/ (x?+2)(x?-2x+1) = 0
& (x? +2)(x-1/=0
<> x? 42 =O hay (x-1)°=0
& x’=-2 (v6 li) hay x- 1=0
©Sx=l
5⁄ 2x(x +3) - 4(x +3) = 0
© (x+3)(2x-4) =0
© x+3=0 hay 2x—4=0
©x=-3 hayx=2
6/ x’ (x+1) - 4(x+1)=0
© (x+1)(x?-4) =0
© (x+l)(x-2)(x+2) =0
©x=-—l hay x= 2hayx=-—2
7/ _(2x—5)(x+4) —-(x+4)(x—3) = 0
© (x+4).[(2x—5)-(x—3)|=0
© (x+4)(2x—-5—x+3) =0
© (x+4)(x—-2) =0
©x=-4hay x=2
8/ (x—5)’ -2(x-5) =0
© (x—5)(x-5}-2(x—5)=0
© (x—5)[(Œ—5)-2] =0
& (x-5)(x-7)=0
© x=5 hay x =7
9/ 5x(x+1)—7(1+x) = 0
& 5x(x+1)—7(x4+1)= 0
€& (x+1)(5x-7) = 0
Ỷ
©x=-lhayx=z
10/ 2x(Sx—-2)-3(2-5x) = 0
© 2x(5x-2)+3(S5x—2)=0
© (5x-2)(2x+3)=0
©Sx=z hay x= >
25/ 4x?-4x+l = (x—3)”
11/ x(x—2)-3x+6 = 0
©x(x-2)-(3x-6) =0
©x(x-2)-3(x—2) = 0
> (x—2)(x—3) = 0
©x=2hayx=3 12/ 6(x+2) -x? +4=(
© 6(x+2) -(x?—4) =0
© 6(x+2) -(x+2)(x—2) =0
© (x+2)|6-(x—2)|=0
© (x+2)(6-x+2) =0
€& (x+2)(8—x)=0
©x=-2hayx=8 13/ x°4+2x*-x-2=0
©(xÌ+2x?)-(x+2)=0
© x”(x+2)-(x+2) =0
& (x+2)(x?-1)=0
€& (x+2)(x—1)(K+1)=0
©&x=-2 hay x=1hayx=-1 14/ x”=2x
© x”-2x=0
<©x(x-2)=0
© x =O hay x-2 =0
©x= Ohayx=2 15/ (3x+2)(x-4)=2x(x-4)
© (3x+2)(x—-4)-2x(x-4) = 0
© (x-4)[(3x+2)-2x] =0
© (x-4)(3x+2-2x) =0
© (x-4)(x+2)=0
& x=4 hay x =-2 16/ 5(x+3)(x—-2)= 3(x+5)(x-2)
© 5(x+3)(x-2)- 3(x+5)(x-2) =0
© (x—-2)|5(x+3)-3(x+5)] =0
> (x—2)(5x+15—3x-15) = 0
€& (x—2)(2x) =0
& x—2 =O hay 2x =0
@x= 2hayx=0 17/ (x-1)(2x—1)=x(1-x)
€& (x—1)(2x—1)—x(1—x) = 0
& (x—1)(2x-1) +x(x-1) = 0
& (x—1)(2x—14+x) =0
& (x—1)(x—1)=0
& x-1=0
©Sx=l 18/ (2x+2)(x-3)= 3(x+l)
© 2(x+l1)(x—-3)= 3(x+l)
© 2(x+l)(x—-3)- 3(x+l)= 0
© (x+l)|2(x—3)-3| =0
© (x+l)(2x-6—3) =0
& (x+1)(2x-9) = 0
©x=-lhayx=4.,5
19/ 3x—-12 = 5x(x-4)
© 3(x-4) = 5x(x-4)
©>3(x-4)- 5x(x-4) = 0Ô
© (x-4)(-5x)=0
3
©x=4hayx=z
5
20/ x”-4=0
© (x-2)(x+2) =0
© x-2 =0 hay x+2 =0
©x= 2hayx=-2 21/ *Cách 1: 16x?=1
© (4x)—I” =0
& (4x-1)(4x+1) =0
& 4x— 1 =O hay 4x+1 =0
©Sx=7 hay x= a 22/ *Cách 1:8x?=2
©x'=sg
1
ex =7
<©>x7=+
*Cách 2: 8x?=2
& 8x’?-2=0
& 2(4x?-1)=0
<> 2(2x—1)(2x+1) =0
© 2x—l =0 hay 2x+l =0
=x - hay x= = 23/ (2x-1)’= 9x’
© (2x-1}—9x”=0
© (2x-IÝ—(3x) =0
© [(2x—I1)-3x|[(2x—1)+3x] =0
© (-x-I)(5x-l)=0
©-x—l =0 hay 5x—l =0
1
©x=-lhayx=
24/ (5x—3)—(4x—7)” = 0)
©I(Sx-3)-(4x—7)]|[(Sx—3)+(4x—7)|=0
© (Sx-3-4x+7)(S5x—-3+4x—7) = 0
& (x+4)(9x-10)=0
© x+4 =0 hay 9x— 10 =0
10
©x=-4hayx= "2
2+3x x-2
———<
4l/ —2 2
Trang 2
<> (2x-1)"-(x-3)"= 0 © 2x— 5x<-—3 +5 © 2(2+3x) < 3(x-—2)
©|(2x—1)-(x-3)][(2x—1)+(x-3)|E0 | ©~3x <2 44 6x < 3x- 6
26/_ (2x+7)”~9(x+2)”= 0 © 142x — 2>3- 2x 42/ ~~ *
©(_x+1)(5x+13) =0 1 43/ ~3x<°
© I6x?+8x +l =0 © x—-x-6x<-3~ 2 ~
—1
29/ x’ +3x-4=0 *Cách 2: (x-3)’ < x”-3 ©3-x<0
© (x?—x)+(4x - 4) =0 © xi-3x-3X+ 2< x?—-3 ©>x>3
©x(x-~ I)+4(x—l)=0 & x°—x*-—3x-3x < —3 -9 A6/ 2 <0
30/ 2x-7 <0 38/ Xxx-3) - =2) <0 4—-2x 13—5x„-—x~3
ĂGx<7 © x -3x⁄— x +4x— 4<0 © 4(1-2x)—1(13-5x) > 3(—x-3)
@xXS3, C> x?~4x44>3~ [4x420- x?—5x| X— — (4-x)(x+ © —-8x + 5x +3x 2>-9-4413
“9 & x?—x?_dx 44x 5x > 3-20-4 | @ 020 (ding)
> —3x >-6 40/ 2(2x-1)? +6 > 8(x+3)(x-3) |~ 2 3 1 ( 6)
exe &> 2(4 x? 4x41) +6 > 8(x?-9) > 3(x—2) —4 > 6(x-1)
@x<2 & 8x?— 8x?_8x >-72 2-6 > 3x — 6x > 6 +644
Giải các phương trình sau:
Trang 3TRUGNG THCS PHU THO QUAN 11 GV: CAO MINH TAI TRANG 3
oes hay 1—2x =—5 * MTC: (x—2)
> -2x =5-1 hay -2x=-5-1 “DKXD: x #2
> -2x=4 hay —-2x=-6 * QD:
©x=-2 hayx=3
(l)«©2x+l =x-2
& 2x—- x =-—2-1
~ (x —2\x +2)
_ (x+(x-3)
©x=-3 (nhận)
Vậy S={-3}
50/ |3x—5|=~4 „ 1
Vi—4 < 0 nên phương trình 56 ` 212v 7
x+2 x-2
51/ [l-2x|~9=-4 * MTC: (x+2)(x—2)
© |1-2x|=5 (cau 49) * QD:
(1) © 1(x—2) — 1(x42) = 3x-12
52/ |2x-5|=3x & x- 2-x -2 =3x- 12
of) 2x-5=3x hay 2x —-5=-3x S&S x= 5 (nhận)
x20
là 3x =5 hay 2x+3x=5 Vay 8)
x20
?Í_x=5 hay 5x=5 57) 3K 3) *
x =—5 (loai) hay x = 1 (nhân) 2-3) 2+9
Vậy S= {4
s3/ |4-3x|+2=x
= |4-3x|=x-2
* MTC: 2(x—3)(x+1)
*PDKXD: x #3;x 4-1
* OD: (1) x(x+l) + x(x—3) =4x
© x?+x+x?-3x-4x=0
©x= 0O(nhận) hay x = 3 (loại)
~ (x(x +1)
*DKXD: x #1;x 4-1
(1) 6 9(x-1) — 1(x41) = 12x
4-3x=x-2 hay 4- 3x=-x+2 & 2x(x — 3) =0
—=3x—x=—2- 4 hay - 3x +x=2—4
of Vay S= {9}
-4x =-6 hay — 2x =-—2 3
> x=Š (loai) hay x =1 (loai) * MTC: 6(x-1)(x+1)
54/ 2|x+1]+1=9
@ 2kx+1=9-1 > 9x —9-x -1 =-12x
4>0 (d > ung) Vậy 3= 12
2x-3
x(2x + 3)
~ x(2x — 3\(2x +3)
* MTC: x(2x+3)(2x—3)
-3 3
“DKXD: x #0; x # > ,X#2
*QD: (1) ©(2x-3Ÿ +(x+l)(2x+3) = x+6
s0/ x+5 _ Xx-5 _ x+5
x-5x 2x?+10x 2x*-50
x+5 x5 x+5
x(x—-5) 2x(x+5) 2(x*-25)
x +5 x—-S x +5
x5) 2x@&«x+5) 2œ&-5œ&+5) (Ủ
* MTC : 2x(x-5)(x+5)
*DKXD:x40;x453;x#-—5
*QÐ: (1) © 2(x45) — (x-5)? = x(x+5)
<> 2( x* +10x+25)—( x? —10x+25) = x” 45x
> 2x? +20x+50— x” +10x —25 — x? -5x =0
& 25x = —25
©x=-Ì (nhận)
Vậy S= [-†
x+†1 x+2 x+3 x14
60/ 99 + 98 = 97 + 96
© (CC +0+(“+09= Sine +1)
x+1+99 x+2+98 x+3+97 x+4+96
x+100 x+100 x+100 x+100
x+100 x+100 x+100 x+100
=0
& (x +100)| — +— —-— 1 ìg
99 98 97 967
©x+100=0 @x=-100 Vậy S= {-100} 1909-x 1907-x 1905-x 1903-x
1909—x 1907—x 1905—x
+1) +( +1) +(
1903—x
2000-x 2000-x 2000-x _ 2000-x
<> (2000 — ¬ `
91 93 95 97 }
©2000-x=0 «x=2000 Vậy S= {2000} x-5 x-15 x-1980 x-1990
62/ 1990 + 1980 = 15 + 5
©œ(——-9+€——-9=(——-9+(——¬"
x-1995 | x-1995 _ x-1995 | x-1995
© (x—1998) "“~“ 1
1990 1980 15 57
©x-1995=0 ©x=1995
Vay S= 11995}