ĐK: \(\left\{{}\begin{matrix}x\ne-1\\y\ge0\\y\ne4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2\left(x+1\right)+1}{x+1}+\frac{4}{\sqrt{y}-2}=4\\\frac{5}{x+1}-\frac{2}{\sqrt{y}-2}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{4}{\sqrt{y}-2}=2\\\frac{5}{x+1}-\frac{2}{\sqrt{y}-2}=4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x+1}=a\\\frac{1}{\sqrt{y}-2}=b\end{matrix}\right.\) \(\left(a,b\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+4b=2\\5a-2b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{x+1}=\frac{10}{11}\\b=\frac{1}{\sqrt{y}-2}=\frac{3}{11}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{10}\\y=\frac{289}{9}\end{matrix}\right.\)
