\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{x-y}=4\left(1\right)\\x^2+y^2=128\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^2+y^2=128\)
\(\Leftrightarrow2x^2+2y^2=256\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2=256\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-y\right)^2=256\)
Hệ phương trình tương đương:
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{x-y}=4\\\left(x+y\right)^2+\left(x-y\right)^2=256\end{matrix}\right.\)(*)
Đặt \(\sqrt{x+y}=a;\sqrt{x-y}=b\), (*) trở thành:
\(\left\{{}\begin{matrix}a+b=4\\a^4+b^4=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left(a^2+b^2\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left(16-2ab\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\256-64ab+4\left(ab\right)^2-2\left(ab\right)^2=256\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\2\left(ab\right)^2-64ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\2ab\left(ab-32\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left[{}\begin{matrix}ab=0\\ab=32\end{matrix}\right.\end{matrix}\right.\)
Tới đây Vieta đảo làm tới thôi :)
