Question

Нуждаюсь в помощи!
x+4>2√4-2x^2
4-2x^2 - это подкоренное выражение

Answer 1

$x+4 textgreater 2sqrt{4-2x^2}, \ left{begin{array}{c} x+4 textgreater 0,\4-2x^2 geq0,\(x+4)^2 textgreater 4(4-2x^2);end{array}right. left{begin{array}{c} x textgreater -4,\x^2-2leq0,\9x^2-8x textgreater 0;end{array}right. left{begin{array}{c} x textgreater -4,\(x+sqrt{2})(x-sqrt{2})leq0,\x(x-frac{8}{9}) textgreater 0;end{array}right. \ left{begin{array}{c} x textgreater -4,\-sqrt{2}leq xleqsqrt{2},\ left [ {{x textless 0,} atop {x textgreater frac{8}{9};}} right. end{array}right. left [ {{-sqrt{2}leq x textless 0,} atop {frac{8}{9} textless xleqsqrt{2};}} right.$
$xin[-sqrt{2};0)cup(frac{8}{9};sqrt{2}]$