Решите неравенство x^2-3|x+1|+2x≤-1.

\[x^{2} - 3|x + 1| + 2x \leq - 1\]

\[x^{2} + 2x + 1 \leq 3|x + 1|\]

\[(x + 1)^{2} \leq 3|x + 1|\]

\[y = |x + 1|:\]

\[y^{2} \leq 3y\]

\[y^{2} - 3y \leq 0\]

\[y(y - 3) \leq 0\]

\[0 \leq y \leq 3.\]

\[0 \leq |x + 1| \leq 3\]

\[0 \leq |x + 1| - при\ всех\ \text{x.}\]

\[|x + 1| \leq 3\]

\[- 3 \leq x + 1 \leq 3\]

\[- 4 \leq x \leq 2\]

\[Ответ:x \in \lbrack - 4;2\rbrack.\]