Question

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

Answer 1

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

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

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

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

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