Решите неравенство: ((x^2-4x)/(x-2))^2+(x^2-4x)/(x-2)-6<=0.

\[\left( \frac{x^{2} - 4x}{x - 2} \right)^{2} + \frac{x^{2} - 4x}{x - 2} - 6 \leq 0\]

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

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

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

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

\[\Longrightarrow - 3 \leq \frac{x^{2} - 4x}{x - 2} \leq 2\]

\[\left\{ \begin{matrix} \frac{x^{2} - 4x}{x - 2} \geq - 3 \\ \frac{x^{2} - 4x}{x - 2} \leq 2\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 4x + 3 \cdot (x - 2)}{x - 2} \geq 0 \\ \frac{x^{2} - 4x - 2 \cdot (x - 2)}{x - 2} \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 4x + 3x - 6}{x - 2} \geq 0\ \\ \frac{x^{2} - 4x - 2x + 4}{x - 2} \leq 0 \\ \end{matrix} \right.\ \]

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

\[\left\{ \begin{matrix} \frac{(x - 3)(x + 2)}{x - 2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{\left( x - 3 - \sqrt{5} \right)\left( x - 3 + \sqrt{5} \right)}{x - 2} \leq 0 \\ \end{matrix} \right.\ \]