\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\)(ĐK: \(x,y\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+\sqrt{y+1}\right)^2=1\\\left(\sqrt{y}+\sqrt{x}+1\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y+1+2\sqrt{x\left(y+1\right)}=1\\y+x+1+2\sqrt{y\left(x+1\right)}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+1+2\sqrt{x\left(y+1\right)}-y-x-1-2\sqrt{y\left(x+1\right)}=0\\x+y+1+2\sqrt{x\left(y+1\right)}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x\left(y+1\right)}-2\sqrt{y\left(x+1\right)}=0\\x+y+1+2\sqrt{x\left(y+1\right)}=1\end{matrix}\right.\)(*)
Giải (*): \(2\sqrt{x\left(y+1\right)}-2\sqrt{y\left(x+1\right)}=0\)
\(\Leftrightarrow2\sqrt{x\left(y+1\right)}=2\sqrt{y\left(x+1\right)}\)
\(\Leftrightarrow\sqrt{x\left(y+1\right)}=\sqrt{y\left(x+1\right)}\)
\(\Leftrightarrow\left(\sqrt{x\left(y+1\right)}\right)^2=\left(\sqrt{y\left(x+1\right)}\right)^2\)
\(\Leftrightarrow x\left(y+1\right)=y\left(x+1\right)\)
\(\Leftrightarrow xy+x=xy+y\)
\(\Leftrightarrow xy+x-xy-y=0\)
\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Ta có HPT: \(\left\{{}\begin{matrix}x=y\\x+y+1+2\sqrt{x\left(y+1\right)}=1\left(1\right)\end{matrix}\right.\)
Thay \(x=y\) vào (1) ta có:
\(y+y+1+2\sqrt{y\left(y+1\right)}=1\)
\(\Leftrightarrow2y+2\sqrt{y^2+y}=0\)
\(\Leftrightarrow2\sqrt{y^2+y}=-2y\)
\(\Leftrightarrow\left(2\sqrt{y^2+y}\right)^2=\left(-2y\right)^2\)
\(\Leftrightarrow4\left(y^2+y\right)=4y\)
\(\Leftrightarrow4y^2+4y=4y\)
\(\Leftrightarrow4y^2=0\Leftrightarrow y^2=0\Leftrightarrow y=0=x\)
Vậy HPT có nghiệm là \(\left(0;0\right)\)
