Решите неравенство: (x^2-4x)^2+7*(x^2-4x)+12<=0.

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

\[t^{2} + 7t + 12 \leq 0\]

\[(t + 3)(t + 4) \leq 0\]

\[- 4 \leq t \leq - 3 \Longrightarrow\]

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

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

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

\[\left\{ \begin{matrix} (x - 2)^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 3)(x - 1) \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\ \ \lbrack 1;3\rbrack.\]