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

murplushka murplushka    1   27.08.2019 07:40    0

Ответы
gxgvxf gxgvxf  05.10.2020 22:24
|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

----------+------[-2/3]----- - -----[1/3]-----+--------
//////////////////////                     //////////////////
----+------[-1]----- - -----[2/9]----------+-----------
///////////////                  //////////////////////////

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