1) ĐK: $x \neq -1$. Ta có
$\dfrac{3(2x-7)}{x+1} = \dfrac{1}{4}$
$<-> 12(2x-7) = x+1$
$<-> 23x = 85$
$<-> x = \dfrac{85}{23}$
2) ĐK: $x \neq -1,3$. Ta có
$\dfrac{x}{2(x-3)} + \dfrac{x}{2(x+1)} - \dfrac{2x}{(x+1)(x-3)} = 0$
$<-> \dfrac{x}{2} \left( \dfrac{1}{x-3} + \dfrac{1}{x+1} - \dfrac{4}{(x+1)(x-3)} \right) = 0$
Vậy $x = 0$ hoặc
$\dfrac{1}{x-3} + \dfrac{1}{x+1} - \dfrac{4}{(x+1)(x-3)} = 0$
$<-> x+1 + x-3 -4 = 0$
$<-> x = 3$ (loại)
Vậy $x = 0$
3) ĐK: $x \neq 0, 2$. Ta có
$\dfrac{2x+2}{x-2} + \dfrac{2}{x(2-x)} = \dfrac{1}{x}$
$<-> \dfrac{x(2x+2)}{x-2} - \dfrac{2}{x(x-2)} - \dfrac{x-2}{x(x-2)} = 0$
$<-> 2x^2 + 2x - 2 - x + 2 = 0$
$<-> 2x^2 +x = 0$
$<-> x(2x + 1) = 0$
$<-> 2x + 1 = 0$ (do $x \neq 0$)
$<-> x = -\dfrac{1}{2}$
Vậy $x = -\dfrac{1}{2}$
