⇔
0}} \right. " alt=" \left \{ {{x+1<(x-1)^2} \atop {x+1 \geq 0}} \atop {x-1>0}} \right. " align="absmiddle" class="latex-formula"> ⇔
1}} \right. " alt=" \left \{ {{x+11}} \right. " align="absmiddle" class="latex-formula"> ⇔
0} \atop {x \geq-1}} \atop {x>1}} \right. " alt=" \left \{ {{x^2-2x+1-x-1>0} \atop {x \geq-1}} \atop {x>1}} \right. " align="absmiddle" class="latex-formula"> ⇔
0} \atop {x \geq-1}} \atop {x>1}} \right. " alt=" \left \{ {{x^2-3x>0} \atop {x \geq-1}} \atop {x>1}} \right. " align="absmiddle" class="latex-formula"> ⇔
0} \atop {x \geq-1}} \atop {x>1}} \right. " alt=" \left \{ {{x(x-3)>0} \atop {x \geq-1}} \atop {x>1}} \right. " align="absmiddle" class="latex-formula">
_____________0\\\\\\\\\\\\\\\\\\\\\_3_\\\\\\\\\\\\\\\\\_x
___________________1/////////////////////////////////_x
______-1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\_x
Ответ: x∈(3;+бесконечность)