\(\left\{{}\begin{matrix}\dfrac{1}{2x+y}+\sqrt{y}=2\\\dfrac{3}{2x+y}+2\sqrt{y}=5\end{matrix}\right.\)
Đặt \(\dfrac{1}{2x+y}=a;\sqrt{y}=b\)
ĐK: x, y ≥ 0
⇒ \(\left\{{}\begin{matrix}a+b=2\\3a+2b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
Trả ẩn: \(\left\{{}\begin{matrix}\dfrac{1}{2x+y}=1\\\sqrt{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x+1}=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy hpt có nghiệm (x ; y) = (0 ; 1)