Question

Тема: иррациональные уравнения

Answer 1

$\sqrt{8-5x}\geq\sqrt{x^{2}-16 }\\\\\left \{ {{x^{2}-16\geq0} \atop {8-5x \geq x^{2} -16 }} \right. \\\\\left \{ {{(x-4)(x+4)\geq0 } \atop {-x^{2}+16+8-5x\geq0}} \right.\\\\\left \{ {{x\in(-\infty;-4]\cup[4;+\infty)} \atop {x^{2}+5x-24\leq0}} \right.\\\\\left \{ {{x\in(-\infty;-4]\cup[4;+\infty)} \atop {(x+8)(x-3)\leq0}} \right.\\\\\left \{ {{x\in(-\infty;-4]\cup[4;+\infty)} \atop {x\in[-8;3]}} \right.$

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______[-8]_____[-4]_____[3]_____[4]______

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Ответ : x ∈ [-8; - 4]