Решите неравенство: x^2+2|x-1|+7<=4|x-2|.

Ответ:

\[x^{2} + 2|x - 1| + 7 \leq 4|x - 2|\]

\[1)\ \left\{ \begin{matrix} x < 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2 - 2x + 7 \leq 8 - 4x \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x < 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x + 1 \leq 0 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 1)^{2} \leq 0 \\ \end{matrix} \right.\ \ ;\ \ \ \ x = - 1.\]

\[2)\ \left\{ \begin{matrix} 1 \leq x \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 2 + 7 \leq 8 - 4x \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 1 \leq x \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 6x - 3 \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} + 6x - 3 = 0\]

\[D = 36 + 12 = 48\]

\[x_{1} = \frac{- 6 \pm 4\sqrt{3}}{2} = - 3 \pm 2\sqrt{3}\]

\[Так\ как\ \ \ \ 1 \leq x \leq 2 \rightarrow x \in \ \varnothing.\]

\[3)\ \left\{ \begin{matrix} x \geq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 2 + 7 \leq 4x - 8 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x \geq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x + 13 \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 2x + 13 = 0\]

\[D = 4 - 52 < 0 \rightarrow x \in \ \varnothing.\]

\[Ответ:\ x = - 1.\]