$Đk;x\ne-1;y\ge0$
$\left\{\begin{matrix}\dfrac{2x}{x+1}+\sqrt{y}=-1\\\dfrac{1}{x+1}+2y=4\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}2x+\sqrt{y}(x+1)=-(x+1)\\1+2y(x+1)=4(x+1)\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}(x+1)(\sqrt{y}+3)=2\\(x+1)(4-2y)=1\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}\sqrt{y}+3=2(4-2y)\\(x+1)(4-2y=1\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}4y+\sqrt{y}-5=0\\(x+1)(4-2y)=1\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}\left[\begin{matrix}\sqrt{y}=\dfrac{-5}{4}\ (loại)\\\sqrt{y}=1\Rightarrow y=1\ (t/m)\end{matrix}\right.\\(x+1)(4-2y)=1\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}y=1\\x=\dfrac{-1}{2}\ (t/m)\end{matrix}\right.$
Vậy $(x;y)=(\dfrac{-1}{2};1)$ là nghiệm của hpt