ĐKXĐ: ..
Đặt \(\left\{{}\begin{matrix}\sqrt{x+y}=a\ge0\\\sqrt{x-y}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{a^2+b^2}{2}\\y=\frac{a^2-b^2}{2}\end{matrix}\right.\)
Ta được hệ:
\(\left\{{}\begin{matrix}a+b=4\\\left(\frac{a^2+b^2}{2}\right)^2+\left(\frac{a^2-b^2}{2}\right)^2=128\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\a^4+b^4=256\end{matrix}\right.\)
Do \(\left\{{}\begin{matrix}a;b\ge0\\a+b=4\end{matrix}\right.\) \(\Rightarrow0\le a;b\le4\Rightarrow0\le a^3;b^3\le64\)
\(\Rightarrow\left\{{}\begin{matrix}a\left(a^3-64\right)\le0\\b\left(b^3-64\right)\le0\end{matrix}\right.\) \(\Leftrightarrow a^4+b^4\le64\left(a+b\right)=256\)
Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}\left(a;b\right)=\left(0;4\right)\\\left(a;b\right)=\left(4;0\right)\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(8;8\right);\left(8;-8\right)\)
