Đặt \(\left\{{}\begin{matrix}x+2=a\\y-1=b\end{matrix}\right.\)
\(\left(a+\sqrt{a^2+1}\right)\left(b+\sqrt{b^2+1}\right)=1\)
\(\Rightarrow\left\{{}\begin{matrix}b+\sqrt{b^2+1}=\sqrt{a^2+1}-a\\a+\sqrt{a^2+1}=\sqrt{b^2+1}-b\end{matrix}\right.\)
\(\Rightarrow a+b+\sqrt{a^2+1}+\sqrt{b^2+1}=\sqrt{a^2+1}+\sqrt{b^2+1}-a-b\)
\(\Rightarrow a+b=0\)
\(\Rightarrow x+2+y-1=0\)
\(\Rightarrow x+y=-1\)
\(\sqrt{x^2+5x+4}\) hay \(\sqrt{x^2+4x+5}\) thế bạn
Lời giải:
ĐKĐB \(\Rightarrow (x+2-\sqrt{x^2+4x+5})(x+2+\sqrt{x^2+4x+5})(y-1+\sqrt{y^2-2y+2})=x+2-\sqrt{x^2+4x+5}\)
\(\Leftrightarrow -(y-1+\sqrt{y^2-2y+2})=x+2-\sqrt{x^2+4x+5}\)
\(\Leftrightarrow \sqrt{x^2+4x+5}-\sqrt{y^2-2y+2}=x+y+1(*)\)
ĐKĐB \(\Rightarrow (x+2+\sqrt{x^2+4x+5})(y-1+\sqrt{y^2-2y+2})(y-1-\sqrt{y^2-2y+2})=y-1-\sqrt{y^2-2y+2}\)
\(\Leftrightarrow -(x+2+\sqrt{x^2+4x+5})=y-1-\sqrt{y^2-2y+2}\)
\(\Leftrightarrow \sqrt{y^2-2y+2}-\sqrt{x^2+4x+5}=x+y+1(**)\)
Lấy $(*)+(**)\Rightarrow x+y+1=0$
$\Leftrightarrow x+y=-1$
