a, x+3 / x-1 - 1 / x-3 + 8 / (x-1)(x-3) = 0
ĐKXĐ x-1 khác 0⇔x khác 1
x-3 khác 0⇔x khác 3
⇔x²-9 / (x-1)(x-3) - (x-1) / (x-1)(x-3) + 8 / (x-1)(x-3) = 0
⇒x²-9-x+1+8=0
⇔x²-x=0
⇔x(x-1)=0
x=0(TM)
⇒
x-1=0⇔x=1(Loại)
Vậy S={0}
b, |2x-1|=3x-9
Vì |2x-1|≥0⇒3x-9≥0⇔x≥3
2x-1=3x-9⇔x=8(TM)
⇒
2x-1=-3x+9⇔5x=10⇔x=2(ko TM)
Vậy S={8}
c,x+1 / x²-5x - 2 / x = 3 / x-5
⇔x+1 / x(x-5) - 2 / x = 3 / x-5
ĐKXĐ x khác 0
x-5 khác 0⇔x khác 5
⇔x+1 / x(x-5) - (2x-10) / x(x-5) = 3x / x(x-5)
⇔x+1-2x+10=3x
⇔-4x=-11
⇔x=11/4(TM)
Vậy S={11/4}
d,|x-2|=2x-1
Vì x-2 ≥0⇒2x-1≥0⇔x=1/2
x-2=2x-1⇔x=-1(ko TM)
⇒
x-2=-2x+1⇔3x=3⇔x=1(TM)
Vậy S={1}
e,2 / x+1 - 1 / x-2 = 3x-11 / (x+1)(x-2)
ĐKXĐ x+1 khác 0⇔x khác -1
x-2 khác 0⇔x khác 2
⇔2x-4 / (x+1)(x-2) - (x+1) / (x+1)(x-2) = 3x-11 / (x+1)(x-2)
⇔2x-4-x-1=3x-11
⇔-2x=-6
⇔x=3(TM)
Vậy S={3}
f,(x+3)(x-4)=(x+3)(4-3x)
⇔(x+3)(x-4)-(x+3)(4-3x)=0
⇔(x+3)(x-4-4-3x)=0
⇔(x+3)(-2x-8)=0
⇔-2(x+3)(x+4)=0
x+3=0⇔x=-3
⇒
x+4=0⇔x=-4
Vậy S={-3;-4}
g,2x-3=4x+7
⇔-2x=10
⇔x=-5
Vậy S={-5}
h,2x(x-3)+6(x-3)=0
⇔2(x+3)(x-3)=0
x+3=0⇔x=-3
⇒
x-3=0⇔x=3
Vậy S={3,-3}
n,x + 2 / x -2 - 6 / x + 2 = x² / x² -4
⇔x + 2 / x -2 -6 / x + 2 = x² / (x-2)(x+2)
ĐKXĐ x-2 khác 0⇔x khác 2
x+2 khác 0⇔x khác -2
⇔x² + 4x + 4 / (x-2)(x+2) - (6x-12) / (x-2)(x+2) = x² / (x-2)(x+2)
⇒x²+4x+4-6x+12=x²
⇔-2x=-16
⇔x=8(TM)
Vậy S={8}
m,5x / x - 3 + 2 / x + 3 = x² + 5x -6 / x² - 9
⇔5x / x - 3 + 2 / x + 3 = x² + 5x -6 / (x-3)(x+3)
ĐKXĐ x-3 khác 0⇔x khác 3
x+3 khác 0⇔x khác -3
⇔5x² + 15x / (x-3)(x+3) + 2x -6 / (x-3)(x+3) = x² + 5x -6 / (x-3)(x+3)
⇒5x²+15x+2x-6=x²+5x-6
⇔4x²+12x=0
⇔4x(x+3)=0
x=0(TM)
⇒
x+3=0⇔x=-3(Ko TM)
Vậy S={-3}
i,x²-16+5x(x-4)=0
⇔x²-16+5x²-20x=0
⇔6x²-20x-16=0
⇔6x²-24x+4x-16=0
⇔6x(x-4)+4(x-4)=0
⇔(x-4)(6x+4)=0
⇔2(x-4)(3x+2)=0
x-4=0⇔x=4
⇒
3x+2=0⇔x=-2/3
Vậy S={4;-2/3}
k,x - 2 / 6 - x/2 = 5 - 2x / 3
⇔x - 2 / 6 -3x /6 = 10 - 4x /6
⇔x-2-3x=10-4x
⇔2x=12
⇔x=6
Vậy S={6}
Xin 5 Sao + CTLHN
