Решите неравенство |x-1|<=(9x^2)/2+2,5xx-1 под модулем

$|x-1| \leq \frac{9x^2}{2} +2.5x$
$|x-1| \leq \frac{9x^2}{2} + \frac{5x}{2}$
$|x-1| \leq \frac{9x^2+5x}{2}$
$\left \{ {{ x-1 \leq \frac{9x^2+5x}{2} } \atop { x-1 \geq - \frac{9x^2+5x}{2} }} \right.$
$\left \{ {{ 2(x-1) \leq {9x^2+5x}} \atop { 2(x-1) \geq - ({9x^2+5x)}} \right.$
$\left \{ {{ 2x-2 \leq {9x^2+5x}} \atop { 2x-2 \geq - {9x^2-5x}} \right.$
$\left \{ {{ 2x-2 {-9x^2-5x \leq 0}} \atop { 2x-2 +{9x^2+5x \geq 0}} \right.$
$\left \{ {{ {-9x^2-3x+2 \leq 0}} \atop {{9x^2+7x-2 \geq 0}} \right.$
$\left \{ {{ {9x^2+3x-2 \geq 0}} \atop {{9x^2+7x-2 \geq 0}} \right.$
$\left \{ {{ {9(x- \frac{1}{3} )(x+ \frac{2}{3}) \geq 0}} \atop {{9(x- \frac{2}{9})(x+1) \geq 0}} \right.$

$9 x^{2} +3x-2=0$
$D=3^3-4*9*(-2)=81$
$x_1= \frac{-3+9}{18} = \frac{1}{3}$
$x_2= \frac{-3-9}{18} =- \frac{2}{3}$

$9 x^{2} +7x-2=0$
$D=7^2-4*9*(-2)=121$
$x_1= \frac{-7+11}{18} = \frac{2}{9}$
$x_1= \frac{-7-11}{18} = -1$

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Ответ: $(-$ ∞ $;-1]$ ∪ $[ \frac{1}{3} ;+$ ∞ $)$