Решите уравнение: x^2/(x-4)+2x/(4-x)=8/(x-4).

\[\frac{x^{2}}{x - 4} + \frac{2x}{4 - x} = \frac{8}{x - 4}\]

\[\frac{x^{2}}{x - 4} - \frac{2x}{x - 4} - \frac{8}{x - 4} = 0\]

\[\frac{x^{2} - 2x - 8}{x - 4} = 0\ \ \ \ \ \ \ \ \ \ \ x \neq 4\]

\[x^{2} - 2x - 8 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 1 \cdot ( - 8) =\]

\[= + 32 = 36\]

\[x_{1} = \frac{2 + \sqrt{36}}{2} = \frac{2 + 6}{2} = \frac{8}{2} =\]

\[= 4\ \ (не\ подходит)\]

\[x_{2} = \frac{2 - \sqrt{36}}{2} = \frac{2 - 6}{2} = \frac{- 4}{2} =\]

\[= - 2.\]

\[Ответ: - 2.\]