Question

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

Answer 1

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

$x^2-7=|3x-7|, \ left [{{ left { {{3x-7<0,} atop {x^2-7=-(3x-7),}} right. } atop { left { {{3x-7 geq 0,} atop {x^2-7=3x-7;}} right. }} right. left [{{ left { {{x< frac{7}{3} ,} atop {x^2+3x-14=0,}} right. } atop { left { {{x geq frac{7}{3} ,} atop {x^2-3x=0;}} right. }} right. \ x^2+3x-14=0, \ D=65, \ x_1= frac{-3- sqrt{65} }{2}<0<frac{7}{3} , x_2= frac{-3+sqrt{65} }{2}>frac{7}{3}, \ x^2-3x=0, \ x(x-3)=0, x_1=0<frac{7}{3}, \ x-3=0, x_2=3;$
$x_1= frac{-3+sqrt{65} }{2}, x_2=3;$

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

Answer 2

Первое и последнее не очень понятно