Решите уравнение: (1-x)/(x^3-3x^2-4x+12)-2/(x+2)(x-3) =1/(x-2).

\[\ \ \ \ \ \ \ \ \ \ \ x \neq 3\]

\[\frac{1 - x - 2(x - 2)}{\left( x^{2} - 4 \right)(x - 3)} =\]

\[= \frac{(x + 2)(x - 3)}{(x^{2} - 4)(x - 3)}\]

\[1 - x - 2x + 4 =\]

\[= x^{2} - 3x + 2x - 6\]

\[x^{2} - x - 6 = 5 - 3x\]

\[x^{2} - x - 6 - 5 + 3x = 0\]

\[x^{2} + 2x - 11 = 0\]

\[D = b^{2} - 4ac =\]

\[= 4 - 4 \cdot 1 \cdot ( - 11) = 4 + 44 =\]

\[= 48\]

\[x_{1} = \frac{- 2 + 4\sqrt{3}}{2} = - 1 + 2\sqrt{3}\]

\[x_{2} = \frac{- 2 - 4\sqrt{3}}{2} = - 1 - 2\sqrt{3}\]

\[Ответ:\ x = - 1 + 2\sqrt{3};\ \ \]

\[x = - 1 - 2\sqrt{3}.\]