ĐKXĐ: \(-1\le x,y\le1\)
\(\hept{\begin{cases}\sqrt{1-x}+\sqrt{1-y}=\sqrt{2}\left(3\right)\\\sqrt{1+x}+\sqrt{1+y}=\sqrt{6}\end{cases}}\)
<=> \(\hept{\begin{cases}1-x+1-y+2\sqrt{\left(1-x\right)\left(1-y\right)}=2\\1+x+1+y+2\sqrt{\left(1+x\right)\left(1+y\right)}=6\end{cases}}\)
<=> \(\hept{\begin{cases}2\sqrt{1-x-y+xy}=x+y\left(1\right)\\2\sqrt{xy+x+y+1}=4-x-y\left(2\right)\end{cases}}\)
Từ (1) và (2) cộng vế theo vế:
\(2\sqrt{xy-x-y+1}+2\sqrt{xy+x+y+1}=4\)
<=>\(\sqrt{xy-x-y+1}+\sqrt{xy+x+y+1}=2\)(đk: - 1 < = x,y < = 1)
<=> \(xy-x-y+1+xy+x+y+1+2\sqrt{\left(1-x^2\right)\left(1-y^2\right)}=4\)
<=> \(2\sqrt{\left(1-x^2\right)\left(1-y^2\right)}=2-2xy\)
<=> \(\sqrt{x^2y^2-x^2-y^2+1}=1-xy\) (đk: xy < = 1)
<=> \(x^2y^2-x^2-y^2+1=x^2y^2-2xy+1\)
<=> \(x^2+y^2-2xy=0\)
<=> \(\left(x-y\right)^2=0\) <=> \(x=y\)
Thay x = y vào pt (3) => \(2\sqrt{1-x}=\sqrt{2}\) (đk: -1 < = x < = 1)
<=> 4(1 - x) = 2 <=> 4 - 4x = 2 <=> 2 = 4x <=> x = 1/2
=> x = y = 1/2 (tm)
