\sqrt{x^{2}- x-2}\\\\\left \{ {{x^{2}+5x>x^{2}-x-2} \atop {x^{2} -x-2\geq0 }} \right.\\\\\left \{ {{5x+x>-2} \atop {(x-2)(x+1)\geq0 }} \right.\\\\\left \{ {{6x>-2} \atop {(x-2)(x+1)\geq 0}} \right.\\\\\left \{ {{x>-\frac{1}{3} } \atop {x\in(-\infty;-1]\cup[2;+\infty)}} \right." alt="\sqrt{x^{2}+5x }>\sqrt{x^{2}- x-2}\\\\\left \{ {{x^{2}+5x>x^{2}-x-2} \atop {x^{2} -x-2\geq0 }} \right.\\\\\left \{ {{5x+x>-2} \atop {(x-2)(x+1)\geq0 }} \right.\\\\\left \{ {{6x>-2} \atop {(x-2)(x+1)\geq 0}} \right.\\\\\left \{ {{x>-\frac{1}{3} } \atop {x\in(-\infty;-1]\cup[2;+\infty)}} \right." align="absmiddle" class="latex-formula">
Окончательный ответ :
x ∈ [2 ; + ∞)