Решите неравенство: ((x^2-3x)/(x-2))^2+(x^2-3x)/(x-2)-6<=0.

\[\left( \frac{x^{2} - 3x}{x - 2} \right)^{2} + \frac{x^{2} - 3x}{x - 2} - 6 \leq 0\]

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

\[t^{2} + t - 6 \leq 0\]

\[(t - 2)(t + 3) \leq 0\]

\[- 3 \leq t \leq 2 \Longrightarrow\]

\[\Longrightarrow - 3 \leq \frac{x^{2} - 3x}{x - 2} \leq 0\]

\[\left\{ \begin{matrix} \frac{x^{2} - 3x}{x - 2} \geq - 3 \\ \frac{x^{2} - 3x}{x - 2} \leq 2\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 3x + 3 \cdot (x - 2)}{x - 2} \geq 0 \\ \frac{x^{2} - 3x - 2 \cdot (x - 2)}{x - 2} \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 6}{x - 2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ \frac{x^{2} - 5x + 4}{x - 2} \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)}{x - 2} \geq 0 \\ \frac{(x - 4)(x - 1)}{x - 2} \leq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\left\lbrack - \sqrt{6};1 \right\rbrack \cup \left\lbrack \sqrt{6};4 \right\rbrack.\]