Question

$sqrt{4-x}-frac{2}{sqrt{4-x}}leq1$

Answer 1

$sqrt{4-x}-frac{2}{sqrt{4-x}}leq1\ sqrt{4-x}=t\ begin{cases} t>0 \ t-frac{2}{t} leq 1 end{cases} => begin{cases} t>0 \ t^2-t-2 leq 0 end{cases} <=>$
$begin{cases} t>0 \ (t+1)(t-2)leq 0 end{cases} <=> begin{cases} t>0 \ -1 leq t leq 2 end {cases} =>$
$0<t leq2$
$sqrt{4-x} leq 2$
$begin{cases} 4-x geq 0 \ 4-x leq 4 end{cases} <=> begin{cases} x leq 4 \ x geq 0 end{cases} => \ x in [0;4]$
Ответ: [0; 4]