Đáp án:
Giải thích các bước giải:
$\frac{3x-1}{x-1}$- $\frac{2x-5}{x+3}$+ $\frac{4}{(x-1)(x+3)}$ =1
⇒$\frac{(x+3)(3x-1)}{(x-1)(x+3)}$- $\frac{(x-1)(2x-5)}{(x+3)(x-1)}$+ $\frac{4}{(x-1)(x+3)}$ =1(x-1)(x+3)
⇔(x+3)(3x-1)-(x-1)(2x-5)+4=1(x-1)(x+3)
⇔(x+3)(3x-1)-(x-1)(2x-5)+4-1(x-1)(x+3)=0
⇔(x-1)(x+3)(3x-1-2x+5+4-1)=0
⇔(x-1)(x+3)(x+7)=0
x-1=0 x=1
⇔x+3=0⇔ x=-3
x+7=0 x=-7