Question

Пожалуйста помагите решить уравнения.

Answer 1

$\sqrt[3]{50+x}+\sqrt{50-x} =10\\50-x\geq 0, x\leq 50\\\sqrt[3]{50+x} =a\\\sqrt{50-x} =b\\a^{3}=50+x\\b^{2}=50-x\\\left \{ {{a^{3}+b^{2} =100} \atop {a+b=10}} \right.\\ \left \{ {{a^{3}+b^{2} =100} \atop {b=10-a}} \right. \\a^{3}+(10-a)^{2}=100\\a^{3}+100-20a+a^{2} =100\\a^{3}+a^{2}-20a=0\\ a(a^{2}+a-20)=0\\a=0; a^{2}+a-20=0\\ a=0; (a+5)(a-4)=0\\a=0; a=-5, a=4\\a=0; 50+x=0; x=-50\\a=-5; 50+x=-125; x=-175\\a=4; 50+x=64; x=14\\x=-175; x=-50; x=14.$