Cần mỗi ý B, C thôi ạ

`b) |x-1| - 3x = 2 (1)`

TH1 `: x>1`

`(1) ⇔ x- 1 - 3x = 2`

`⇔ x = -3/2` (loại)

TH2 `: x<1`

`(1) ⇔ 1- x - 3x = 2`

`⇔ x = -1/4`

Vậy `S={-1/4}`

`c)` $\dfrac{x+2}{x-2}$ `+` $\dfrac{1}{x}$ `=` $\dfrac{-2}{2x-x²}$

`⇔` $\dfrac{x²+2x}{x²-2x}$ `+` $\dfrac{x-2}{x²-2x}$ `=` $\dfrac{2}{x²-2x}$

`⇔ x²+2x + x - 2 = 2`

`⇔ x² + 3x - 4 = 0`

`⇔` $\left[\begin{matrix} x=1\\ x=-4\end{matrix}\right.$

Vậy `S={1;-4}`