\(\left\{{}\begin{matrix}x+\sqrt{y-1}=1\\y+\sqrt{x-1}=1\end{matrix}\right.\left(1\right)\left(ĐK:x-1\ge0;y-1\ge0\right)\\ \Leftrightarrow\left\{{}\begin{matrix}x-1+\sqrt{y-1}=0\\y-1+\sqrt{x-1}=0\end{matrix}\right.\)
Do x-1\(\ge\)0, y-1\(\ge\)0 nên :
\(\left\{{}\begin{matrix}x-1+\sqrt{y-1}=0\\y-1+\sqrt{x-1}=0\end{matrix}\right.=0\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) (Thỏa ĐK)
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