Đáp án:
$\begin{array}{l}
a)A = \dfrac{{x + \sqrt {{x^2} - 2x} }}{{x - \sqrt {{x^2} - 2x} }} - \dfrac{{x - \sqrt {{x^2} - 2x} }}{{x + \sqrt {{x^2} - 2x} }}\\
Dkxd:\left\{ \begin{array}{l}
{x^2} - 2x \ge 0\\
x - \sqrt {{x^2} - 2x} \# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 2} \right) \ge 0\\
x\# \sqrt {{x^2} - 2x}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
{x^2}\# {x^2} - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
x\# 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
Vậy\,x < 0\,hoặc\,x \ge 2\\
b)A = \dfrac{{x + \sqrt {{x^2} - 2x} }}{{x - \sqrt {{x^2} - 2x} }} - \dfrac{{x - \sqrt {{x^2} - 2x} }}{{x + \sqrt {{x^2} - 2x} }}\\
= \dfrac{{{{\left( {x + \sqrt {{x^2} - 2x} } \right)}^2} - {{\left( {x - \sqrt {{x^2} - 2x} } \right)}^2}}}{{\left( {x + \sqrt {{x^2} - 2x} } \right).\left( {x - \sqrt {{x^2} - 2x} } \right)}}\\
= \dfrac{{{x^2} + 2.x.\sqrt {{x^2} - 2x} + {x^2} - 2x - \left( {{x^2} - 2x\sqrt {{x^2} - 2x} + {x^2} - 2x} \right)}}{{{x^2} - \left( {{x^2} - 2x} \right)}}\\
= \dfrac{{4x\sqrt {{x^2} - 2x} }}{{2x}}\\
= 2\sqrt {{x^2} - 2x} \\
c)A < 2\\
\Leftrightarrow 2\sqrt {{x^2} - 2x} < 2\\
\Leftrightarrow \sqrt {{x^2} - 2x} < 1\\
\Leftrightarrow {x^2} - 2x < 1\\
\Leftrightarrow {x^2} - 2x + 1 < 2\\
\Leftrightarrow {\left( {x - 1} \right)^2} < 2\\
\Leftrightarrow - \sqrt 2 < x - 1 < \sqrt 2 \\
\Leftrightarrow 1 - \sqrt 2 < x < 1 + \sqrt 2 \\
Do:\left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
1 - \sqrt 2 < x < 0
\end{array} \right.\\
Vậy\,1 - \sqrt 2 < x < 0\,hoặc\,2 \le x < 1 + \sqrt 2
\end{array}$
