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

[latex] |x-1| \leq \frac{9x^2}{2} +2.5x[/latex] [latex] |x-1| \leq \frac{9x^2}{2} + \frac{5x}{2} [/latex] [latex] |x-1| \leq \frac{9x^2+5x}{2} [/latex] [latex] \left \{ {{ x-1 \leq \frac{9x^2+5x}{2} } \atop { x-1 \geq - \frac{9x^2+5x}{2} }} \right. [/latex] [latex] \left \{ {{ 2(x-1) \leq {9x^2+5x}} \atop { 2(x-1) \geq - ({9x^2+5x)}} \right. [/latex] [latex] \left \{ {{ 2x-2 \leq {9x^2+5x}} \atop { 2x-2 \geq - {9x^2-5x}} \right. [/latex] [latex] \left \{ {{ 2x-2 {-9x^2-5x \leq 0}} \atop { 2x-2 +{9x^2+5x \geq 0}} \right. [/latex] [latex] \left \{ {{ {-9x^2-3x+2 \leq 0}} \atop {{9x^2+7x-2 \geq 0}} \right. [/latex] [latex] \left \{ {{ {9x^2+3x-2 \geq 0}} \atop {{9x^2+7x-2 \geq 0}} \right. [/latex] [latex] \left \{ {{ {9(x- \frac{1}{3} )(x+ \frac{2}{3}) \geq 0}} \atop {{9(x- \frac{2}{9})(x+1) \geq 0}} \right. [/latex] [latex]9 x^{2} +3x-2=0[/latex] [latex]D=3^3-4*9*(-2)=81[/latex] [latex]x_1= \frac{-3+9}{18} = \frac{1}{3} [/latex] [latex]x_2= \frac{-3-9}{18} =- \frac{2}{3} [/latex] [latex]9 x^{2} +7x-2=0[/latex] [latex]D=7^2-4*9*(-2)=121[/latex] [latex]x_1= \frac{-7+11}{18} = \frac{2}{9} [/latex] [latex]x_1= \frac{-7-11}{18} = -1[/latex] ----------+------[-2/3]----- - -----[1/3]-----+-------- //////////////////////                     ////////////////// ----+------[-1]----- - -----[2/9]----------+----------- ///////////////                  ////////////////////////// Ответ: [latex](-[/latex] ∞ [latex];-1][/latex] ∪ [latex][ \frac{1}{3} ;+[/latex] ∞ [latex])[/latex]