\(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1\)
\(\Leftrightarrow\sqrt{x^2+1}+x=\sqrt{y^2+1}+y\) (1)
Tương tự ta có: \(\sqrt{y^2+1}-y=\sqrt{x^2+1}-x\) (2)
Cộng vế (1) và (2) \(\Rightarrow x-y=y-x\Rightarrow x=y\)
Thế xuống dưới:
\(3\sqrt{3x-2}+x\sqrt{6-x}=10\)
Đặt \(\sqrt{6-x}=a\Rightarrow\left\{{}\begin{matrix}0\le a\le\frac{4\sqrt{3}}{3}\\x=6-a^2\end{matrix}\right.\)
\(\Rightarrow a^3-6a+10-3\sqrt{16-3a^2}=0\)
\(\Leftrightarrow\left(a^3-3a-2\right)+3\left(4-a-\sqrt{16-3a^2}\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+1\right)^2+\frac{12a\left(a-2\right)}{4-a+\sqrt{16-a^2}}=0\)
\(\Leftrightarrow\left(a-2\right)\left[\left(a+1\right)^2+\frac{12a}{4-a+\sqrt{16-a^2}}\right]=0\)
\(\Leftrightarrow a=2\Leftrightarrow...\)
