Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Victorique de Blois]

đk : x;y;z khác 0

A =\(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}=\sqrt[3]{\frac{ax^3}{x}+\frac{ax^3}{y}+\frac{ax^3}{z}}=\sqrt[3]{ax^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)

mà 1/x + 1/y + 1/z = 1 nên A = \(\sqrt[3]{ax^3}=x\sqrt[3]{a}\) => \(\sqrt[3]{a}=\frac{A}{x}\)

tương tự cm đc \(\sqrt[3]{b}=\frac{A}{x}=\sqrt[3]{c}\) 

=> \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=A\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=A\)