Question

Решите неравенство, пожалуйста!(((

Answer 1

$\sqrt{x+1}=x-1$

ОДЗ:
$x+1\ \geq \ 0\\ x\ \geq \ -1$

$\sqrt {x+1}\ \textgreater \ x-1\\ \sqrt {(x+1)^2}\ \textgreater \ (x-1)^2\\ x+1\ \textgreater \ x^2-2x+1\\ x^2-2x-x+1-1\ \textgreater \ 0\\ x^2-3x\ \textgreater \ 0\\\\ x\ \textgreater \ 0\\\\ x-3\ \textgreater \ 0\\ x\ \textgreater \ 3$

Ответ: $x \in [-1;3)$