ĐKXĐ: \(\begin{cases}x^2-1\ge0\\ x^2<>x^2-1\end{cases}\Rightarrow x^2-1\ge0\)
=>\(\left[\begin{array}{l}x\ge1\\ x\le-1\end{array}\right.\)
\(\frac{1}{x-\sqrt{x^2-1}}=\frac{x+\sqrt{x^2-1}}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2+1}\right)}=\frac{x+\sqrt{x^2-1}}{x^2-\left(x^2-1\right)}=x+\sqrt{x^2-1}\)
Ta có: \(x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20\)
=>\(x+\sqrt{x^2-1}+x+\sqrt{x^2-1}=20\)
=>\(2x+\sqrt{x^2-1}=20\)
=>\(\sqrt{x^2-1}=20-2x\)
=>\(\begin{cases}20-2x\ge0\\ \left(20-2x\right)^2=x^2-1\end{cases}\Rightarrow\begin{cases}x\le10\\ 4x^2-80x+400-x^2+1=0\end{cases}\)
=>\(\begin{cases}x\le10\\ 3x^2-80x+401=0\end{cases}\)
\(3x^2-80x+401=0\)
\(\Delta=\left(-80\right)^2-4\cdot3\cdot401=6400-12\cdot401=1588>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{80-\sqrt{1588}}{2\cdot3}=\frac{80-2\sqrt{397}}{6}=\frac{40-\sqrt{397}}{3}\left(nhận\right)\\ x=\frac{80+\sqrt{1588}}{2\cdot3}=\frac{80+2\sqrt{397}}{6}=\frac{40+\sqrt{397}}{3}\left(loại\right)\end{array}\right.\)
