Giải thích các bước giải:
ĐKXĐ:
\(\left\{ \begin{array}{l}
x - \frac{1}{x} \ge 0\\
1 - \frac{1}{x} \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2} - 1}}{x} \ge 0\\
\frac{{x - 1}}{x} \ge 0
\end{array} \right. \Leftrightarrow x \ge 1\)
Ta có:
\(\begin{array}{l}
\sqrt {x - \frac{1}{x}} - \sqrt {1 - \frac{1}{x}} > \frac{{x - 1}}{x}\\
\Leftrightarrow \sqrt {\frac{{{x^2} - 1}}{x}} - \sqrt {\frac{{x - 1}}{x}} > \frac{{x - 1}}{x}\\
\Leftrightarrow \sqrt {\frac{{x - 1}}{x}} \left( {\sqrt {x + 1} - 1} \right) > \frac{{x - 1}}{x}\\
\Leftrightarrow \sqrt {x + 1} - 1 > \sqrt {\frac{{x - 1}}{x}} \\
\Leftrightarrow \sqrt {x\left( {x + 1} \right)} - \sqrt x > \sqrt {x - 1} \\
\Leftrightarrow {x^2} + x - 2x.\sqrt {x + 1} + x > x - 1\\
\Leftrightarrow {x^2} + x + 1 - 2x\sqrt {x + 1} > 0\\
\Leftrightarrow {x^2} - 2x.\sqrt {x + 1} + \left( {x + 1} \right) > 0\\
\Leftrightarrow {\left( {x - \sqrt {x + 1} } \right)^2} > 0\\
\Leftrightarrow x \ne \sqrt {x + 1} \\
\Leftrightarrow {x^2} - x - 1 \ne 0\\
\Leftrightarrow x \ne \frac{{1 \pm \sqrt 5 }}{2}
\end{array}\)
