Đáp án:
`S=∅`
Giải thích các bước giải:
`2/(x^2-4) - (x-1)/(x(x-2)) + (x-4)/(x(x+2)) (\text{ĐK}: x`$\neq0; x $$\neq±2)$
$\Leftrightarrow\dfrac{2x\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(4+x\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0$
`<=>` $\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x-1\right)}-\dfrac{4+x}{x\left(x+2\right)}=0$
$\Rightarrow2x^2-2x-x^2+4-x^2+2x+8=0$
$\Leftrightarrow12=0$ (vô lý)
`=> S=∅`
