$1 - \frac{1}{2 - x} = \frac{6 - x}{3x^2 - 12} - \frac{1}{x - 2}$
$1+ \frac{1}{x - 2} = \frac{6 - x}{3(x^2 - 4)} - \frac{1}{x - 2}$
$1 = \frac{6 - x}{3(x^2 - 4)} - \frac{2}{x - 2}$ |*3(x − 2)(x + 2)
$\begin{equation*} \begin{cases} 3(x - 2)(x + 2) = 6 - x - 2 * 3(x + 2) &\\ x ≠ ±2 & \end{cases} \end{equation*}$
$3(x^2 - 4) = 6 - x - 6x - 12$
$3x^2 + 7x - 6 = 0$
(3x − 2)(x + 3) = 0
3x − 2 = 0
3x = 2
$x_1 = \frac{2}{3}$
x + 3 = 0
$x_2 = -3$
$\begin{equation*} \begin{cases} x_1 = \frac{2}{3}, x_2 = -3 &\\ x ≠ ±2 & \end{cases} \end{equation*}$
Ответ:
$x_1 = \frac{2}{3}$;
$x_2 = -3$.