Đáp án:
$ĐK: x\neq 0 ;x\neq 1 ; x\neq 2 $
$\frac{1}{x-1}$ + $\frac{1}{(x-1)(x-2)}$ =$\frac{3}{2x}$
⇔ $\frac{2x(x-2)}{(x-1)(x-2)}$ + $\frac{2x}{(x-1)(x-2)}$ =$\frac{3(x-1)(x-2)}{2x(x-1)(x-2)}$
⇒ $2x(x-2) + 2x = 3(x-1)(x-2) $
⇔ $2x²-4x+2x=(3x-3)(x-2)$
⇔ $2x²-2x=3x²-6x-3x+6$
⇔ $2x²-2x-3x²+6x+3x-6=0$
⇔ $-x²+7x-6=0$
Ta có : $a+b+c=-1+7-6=0$
⇒ $Pt $ có 2 nghiệm phân biệt :
+ $x_1 = 1 (loại)$
+ $x_2 = c/a = -6 (tm)$
