Giải thích các bước giải:
$DKXD : x\ne 2$
Ta có :
$3x-\dfrac{1}{x-2}=\dfrac{x-1}{2-x}$
$\to 3x\left(x-2\right)\left(-x+2\right)-\dfrac{1}{x-2}\left(x-2\right)\left(-x+2\right)=\dfrac{x-1}{2-x}\left(x-2\right)\left(-x+2\right)$
$\to 3x\left(x-2\right)\left(-x+2\right)-\left(-x+2\right)=\left(x-1\right)\left(x-2\right)$
$\to -3x^3+11x^2-8x-4=0$
$\to -\left(x-2\right)^2\left(3x+1\right)=0$
$\to x\in\{2,\dfrac{-1}3\}$
$\to x=-\dfrac 13$ vì $x\ne 2$
