a, ĐK: \(x\ge1\)
\(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{x^2-1}}+\sqrt{2x+2\sqrt{x^2-1}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}}+\sqrt{x-1+x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-\sqrt{x+1}\right)^2}+\sqrt{\left(\sqrt{x-1}+\sqrt{x+1}\right)^2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+1}-\sqrt{x-1}+\sqrt{x-1}+\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow2\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow x+1=2\)
\(\Leftrightarrow x=1\left(tm\right)\)
b, ĐK: \(x\ge-1+\sqrt{2},x\le-1-\sqrt{2}\)
Đặt \(\sqrt{x^2+2x-1}=t\left(t\ge0\right)\)
\(pt\Leftrightarrow2\left(1-x\right)t=t^2-4x\)
\(\Leftrightarrow t^2-4x+2xt-2t=0\)
\(\Leftrightarrow\left(t-2\right)\left(2x+t\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\\sqrt{x^2+2x-1}=-2x\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-1\pm\sqrt{6}\left(tm\right)\)
