Решите неравенство: |x^2-2x|>=3.

\[\left| x^{2} - 2x \right| \geq 3\]

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

\[\left\{ \begin{matrix} x² - 2x + 3 \leq 0 \\ x² - 2x - 3 \geq 0 \\ \end{matrix} \right.\ \]

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

\[D = 4 - 12 < 0 - x \in R.\]

\[x^{2} - 2x - 3 \geq 0\]

\[x_{1} + x_{2} = 2;\ \ x_{1} \cdot x_{2} = - 3\]

\[x_{1} = 3;\ \ x_{2} = - 1\]

\[(x + 1)(x - 3) \geq 0\]

\[x \leq - 1;\ \ x \geq 3.\]

\[Ответ:( - \infty;\ - 1\rbrack \cup \lbrack 3; + \infty).\]