\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
a) A có nghĩa <=> \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)
b) Nếu \(x\ge\sqrt{2}\)khi đó \(\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\)
Ta có: \(A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)