Question

Giải hệ phương trình \(\hept{\begin{cases}\sqrt{x+y}+\sqrt{xy}=4\\\sqrt{x}+\sqrt{y}=2\sqrt{2}\end{cases}}\)

Answer 1

Đặt \(\sqrt{x}+\sqrt{y}=a,\sqrt{xy}=b\) . Hệ trở thành :

\(\left\{{}\begin{matrix}\sqrt{a^2-2b}+b=4\\a=2\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\sqrt{2}\\8-2b=b^2-8b+16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\sqrt{2}\\b^2-6b+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\sqrt{2}\\b=2\end{matrix}\right.\\\left\{{}\begin{matrix}a=2\sqrt{2}\\b=4\end{matrix}\right.\end{matrix}\right.\)

Tiếp nha bạn :))