Question

(x+2)^2=2|x+2|+3 как делать |3x-2|x< 1

Answer 1

$(x+2)^2=2|x+2|+3\\\\ (x+2)^2=|x+2|^2\\\\ |x+2|^2=2|x+2|+3\\\\|x+2|=t\\\\ t^2=2t+3\\\\ t^2-2t-3=0\\\\ D=4+12=16; \ \sqrt D=4\\\\ t_{1/2}= \frac{2\pm4}{2}\\\\ t_1=3;\ t_2=-1\\\\ |x+2|=3\\\\ \left \{ {{x+2=3} \atop {x+2=-3}} \right. \ \ \left \{ {{x=1} \atop {x=-50}} \right. \\\\\\ |x+2|=-1$
Нет решений

ответ:  $x_1=1; x_2=-5$

$|3x-2|\ \textless \ 1\\\\ \left \{ {{3x-2\ \textless \ 1} \atop {3x-2\ \textgreater \ -1}} \right. \ \ \left \{ {{3x\ \textless \ 3} \atop {3x\ \textgreater \ 1}} \right. \ \ \left \{ {{x\ \textless \ 1} \atop {x\ \textgreater \ \frac{1}{3} }} \right.$

ответ:  $x\in( \frac{1}{3}; 3)$