Question

Срочно помогите решить

Answer 1

$\left \{ {{x^2+x-6\geq 0} \atop {x(x-1)-(x+1)^2\leq 8}} \right.\\D=1+24=5^2\\\left \{ {{(x-\frac{-1-5}{2} )(x-\frac{-1+5}{2} )\geq 0} \atop {x^2-x-x^2-2x-1-8\leq 0}} \right.\\\left \{ {{(x+3)(x-2)\geq 0} \atop {x\geq -3}} \right.$

Ответ: x∈{-3}∪[2;+∞).