Ответ:

х₁ = -2; х₂ = 1

Пошаговое объяснение:
|х+2|*(-х+1) = 0
|х+2|*(1-х) = 0

а) Пусть x+2<0, тогда
[tex]\displaystyle \left \{ {{-(x+2)*(1-x) = 0|:(-1)} \atop {x+2 < 0}} \right. < = > \left \{ {{(x+2)*(1-x) = 0} \atop {x < -2}} \right. < = > \left \{ {{\left[\begin{array}{ccc}x+2=0\\1-x=0\\\end{array}\right} \atop {x < -2}} \right. < = > \left \{ {{\left[\begin{array}{ccc}x=-2\\x=1\\\end{array}\right} \atop {x < -2}} \right.[/tex]
Здесь нет решений
б) Пусть x+2≥0, тогда
[tex]\displaystyle \left \{ {{(x+2)*(1-x) = 0} \atop {x+2 \geq 0}} \right. < = > \left \{ {{\left[\begin{array}{ccc}x+2=0\\1-x=0\\\end{array}\right} \atop {x \geq -2}} \right. < = > \left \{ {{\left[\begin{array}{ccc}x=-2\\x=1\\\end{array}\right} \atop {x \geq -2}} \right.[/tex]
Решение: х₁ = -2; х₂ = 1