\(\left\{{}\begin{matrix}x+y=8\\2x+y+\sqrt{x+1}=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=8-x\\2x+8-x+\sqrt{x+1}=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=8-x\\x-5+\sqrt{x+1}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=8-x\\\left(\sqrt{x+1}-2\right)\left(\sqrt{x+1}+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=8-x\\\sqrt{x+1}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(3;5\right)\)
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\x+x+y+\sqrt{x+1}=13\end{matrix}\right.\) \(\Rightarrow x+8+\sqrt{x+1}=13\)
\(\Leftrightarrow x+1+\sqrt{x+1}-6=0\)
Đặt \(\sqrt{x+1}=a\ge0\Rightarrow a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}=2\Rightarrow x=3\Rightarrow y=5\)
