Question

Решить , уравнения с модулями [2x-3]=3-2x x^2-7=[3x-7] 2[x^2+2x-5]=x-1

Answer 1

$|2x-3|=3-2x, \\ \left \{ {{3-2x \geq 0,} \atop { \left [ {{2x-3=3-2x,} \atop {2x-3=-(3-2x),}} \right. }} \right. \left \{ {{-2x \geq -3,} \atop { \left [ {{4x=6,} \atop {0\cdot x=0,}} \right. }} \right. \left \{ {{x \leq 1,5,} \atop { \left [ {{x=1,5,} \atop {x\in R,}} \right. }} \right. \\ x \leq 1,5, \\ x\in(-\infty;1,5].$

$x^2-7=|3x-7|, \\ \left [{{ \left \{ {{3x-7$
$x_1= \frac{-3+\sqrt{65} }{2}, x_2=3; \\$

$2|x^2+2x-5|=x-1, \\ \left \{ {{x-1 \geq 0,} \atop { \left [ {{2(x^2+2x-5)=x-1,} \atop {2(x^2+2x-5)=-(x-1),}} \right. }} \right. \left \{ {{x \geq 1,} \atop { \left [ {{2x^2+4x-10=x-1,} \atop {2x^2+4x-10=1-x,}} \right. }} \right. \left \{ {{x \geq 1,} \atop { \left [ {{2x^2+3x-9=0,} \atop {2x^2+5x-11=0,}} \right. }} \right. \\ 2x^2+3x-9=0,\\ D=81, \\ x_1=-3$
$x_1=1,5, x_2= \frac{-5+\sqrt{113} }{4}.$