Question

Решить неравенство корень из 5-2x меньше или равно 1-x

Answer 1

$sqrt{5-2x} leq 1-x\ ===========\ left { {{5-2x geq 0} atop {1-x geq 0}} right. \ left { {{-2x geq -5} atop {-x geq -1}} right. \ left { {{x leq 2.5} atop {x leq 1}} right. \ xin (-infty;1] \ ===========\ ( sqrt{5-2x})^2 leq (1-x)^2\ 5-2x leq 1-2x+x^2\ 5-2x-1+2x-x^2 leq 0\ 4-x^2 leq 0\ (2-x)(2+x) leq 0\\ left { {{(2-x)(2+x) leq 0} atop {x leq 1}} right. \\ xin (-infty;-2]$