Question

Cho P=\(P=\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}\). Chứng minh rằng: \(\sqrt[3]{P^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)

Answer 1

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x^2}=a\ge0\\\sqrt[3]{y^2}=b\ge0\end{matrix}\right.\)

\(P=\sqrt{a^3+a^2b}+\sqrt{b^3+ab^2}=\sqrt{a^2\left(a+b\right)}+\sqrt{b^2\left(a+b\right)}\)

\(=a\sqrt{a+b}+b\sqrt{a+b}=\left(a+b\right)\sqrt{a+b}\)

\(\Rightarrow P^2=\left(a+b\right)^2\left(a+b\right)=\left(a+b\right)^3\)

\(\Rightarrow\sqrt[3]{P^2}=a+b=\sqrt[3]{x^2}+\sqrt[3]{y^2}\) (đpcm)