Решите неравенство (x-1)(2x^2-3x+1)(x+5)≥0.

\[(x - 1)\left( 2x^{2} - 3x + 1 \right)(x + 5) \geq 0\]

\[2x^{2} - 3x + 1 = 2 \cdot (x - 0,5)(x - 1)\]

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{3 + 1}{4} = 1;x_{2} = \frac{3 - 1}{4} = 0,5\]

\[(x + 5)(x - 0,5)(x - 1)(x - 1) \geq 0\]

\[Ответ:x \in ( - \infty; - 5\rbrack \cup \lbrack 0,5; + \infty).\]