(x+√(x^2-2x))/(x-√(x^2-2x)) – (x-√(x^2-2x))/(x+√(x^2-2x)) a/tìm ĐKXD b/Rút gọn A c/Tìm x để A<2

Đáp án:

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