Question

Помогите с алгеброй 21 балл

Answer 1

$\sqrt[5]{x} +2\sqrt[10]{x} -8\leq 0\\(a+b)^2=a^2+2ab+b^2\\\\\sqrt[5]{x} +2\sqrt[10]{x} +1-9\leq 0\\(\sqrt[10]{x} +1)^2-9\leq 0\\a^2-b^2=(a-b)(a+b)\\(\sqrt[10]{x} +1-3)(\sqrt[10]{x} +1+3)\leq 0\\\left \{ {{\sqrt[10]{x}+1-3 \leq 0} \atop {x>0}} \right. \\\left \{ {{\sqrt[10]{x} \leq 2} \atop {x>0}} \right. \\\\0<x\leq 2^{10}$