Đk: \(x\ge y\) và \(x+y\ge0\)
\(Pt\left(1\right)\Rightarrow x+y+x-y-2\sqrt{x^2-y^2}=4\)
\(\Leftrightarrow x-2=\sqrt{x^2-y^2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x+4=x^2-y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\y^2=4x-4\end{matrix}\right.\)
\(Pt\left(2\right)\Rightarrow x^2+y^2+1+x^2-y^2-2\sqrt{\left(x^2+y^2+1\right)\left(x^2-y^2\right)}=9\)
\(\Leftrightarrow x^2-4=\sqrt{\left(x^2+y^2+1\right)\left(x^2-y^2\right)}\)
Thay \(y^2=4x-4\) \(\left(x\ge2\right)\)vào pt (2) được:
\(x^2-4=\sqrt{\left(x^2+4x-4\right)\left(x^2-4x+4\right)}\)\(\Leftrightarrow x^2-4=\sqrt{\left(x-2\right)^2\left(x^2+4x-4\right)}\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\sqrt{x^2+4x-4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=\sqrt{x^2+4x-4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x^2+4x+4=x^2+4x-4\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=2\)
\(\Rightarrow y^2=4\) \(\Rightarrow\left[{}\begin{matrix}y=2\left(tm\right)\\y=-2\left(ktm\right)\end{matrix}\right.\)
Vậy (x;y)=(2;2)
