Question

решите неравенства ^2-х>-5

Answer 1

$sqrt{2-x} >-5 \ 2-x geq 0 \ -x geq -2 \ x leq 2$
x∈$(-infty ;2)$
$sqrt{2-x} >-5 \ ( sqrt{2-x})^{2} >(-5 )^{2} \ 2-x>25 \ -x>23 \ x<23$