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

\[\frac{x^{2}}{x - 1} - \frac{3x}{1 - x} = \frac{4}{x - 1}\]

\[\frac{x^{2}}{x - 1} + \frac{3x}{x - 1} - \frac{4}{x - 1} = 0\]

\[\frac{x^{2} + 3x - 4}{x - 1} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 1\]

\[x^{2} + 3x - 4 = 0\]

\[D = 3^{2} - 4 \cdot 1 \cdot ( - 4) = 9 + 16 =\]

\[= 25\]

\[x_{1} = \frac{- 3 + \sqrt{25}}{2} = \frac{- 3 + 5}{2} = \frac{2}{2} =\]

\[= 1\ (не\ подходит)\]

\[x_{2} = \frac{- 3 - \sqrt{25}}{2} = \frac{- 3 - 5}{2} =\]

\[= \frac{- 8}{3} = - 4.\]

\[Ответ:\ x = - 4.\]