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

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

\frac{7}{3}, \\ x^2-3x=0, \\ x(x-3)=0, x_1=0<\frac{7}{3}, \\ x-3=0, x_2=3; \\" alt="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; \\" align="absmiddle" class="latex-formula">
$x_1= \frac{-3+\sqrt{65} }{2}, x_2=3; \\$

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

arsenlevadniy_zn (93.5k баллов) 27 Март, 18