Question

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2=-3\\\sqrt[4]{x}+\sqrt{32-x}+6y=24\end{matrix}\right.\)

Answer 1

Cộng vế:

\(\sqrt{x}+\sqrt[]{32-x}+\sqrt[4]{x}+\sqrt[4]{32-x}=y^2-6y+21\)

\(\Leftrightarrow\sqrt[]{x}+\sqrt[]{32-x}+\sqrt[4]{x}+\sqrt[4]{32-x}=\left(y-3\right)^2+12\)

Ta có:

\(\sqrt{x}+\sqrt[]{32-x}\le\sqrt{2\left(x+32-x\right)}=8\)

\(\sqrt[4]{x}+\sqrt[4]{32-x}\le\sqrt{2\left(\sqrt[]{x}+\sqrt[]{32-x}\right)}\le\sqrt{2.8}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt[]{x}+\sqrt[]{32-x}+\sqrt[4]{x}+\sqrt[4]{32-x}\le12\\\left(y-3\right)^2+12\ge12\end{matrix}\right.\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x=32-x\\y=3\end{matrix}\right.\)