ĐK: \(x,y\ge2\)
\(\left\{{}\begin{matrix}\sqrt{x+1}+\sqrt{y-2}=3\\\sqrt{y+1}+\sqrt{x-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}+\sqrt{y-2}=3\\\sqrt{y+1}+\sqrt{x-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{\left(x+1\right)\left(y-2\right)}=10\left(1\right)\\x+y+2\sqrt{\left(y+1\right)\left(x-2\right)}=10\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x+1\right)\left(y-2\right)}=\sqrt{\left(y+1\right)\left(x-2\right)}\)
\(\Leftrightarrow xy-2x+y-2=xy-2y+x-2\)
\(\Leftrightarrow x=y\)
\(\left(1\right)\Leftrightarrow2x+2\sqrt{\left(x+1\right)\left(x-2\right)}=10\)
\(\Leftrightarrow\sqrt{x^2-x-2}=5-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x-2=x^2-10x+25\\5-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x=27\\x\le5\end{matrix}\right.\Leftrightarrow x=y=3\left(tm\right)\)
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