Đáp án:

a)ĐKXĐ:\(\begin{cases}x^2-2x \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} \ne 0\\\end{cases}\)

`<=>` \(\begin{cases}x(x-2) \ge 0\\x \ne 0\\x \ne 0\\\end{cases}\)

`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge2\\x \le 0\end{array} \right.\\x \ne 0\\\end{cases}\)

`<=>` \(\left[ \begin{array}{l}x\ge 2\\x<0\end{array} \right.\)

`b)A=(x+\sqrt{x^2-2x})/(x-\sqrt{x^2-2x})-(x-\sqrt{x^2-2x})/(x+\sqrt{x^2-2x})`

`=(x+\sqrt{x^2-2x})^2/(x^2-(x^2-2x))-((x-\sqrt{x^2-2x})^2)/(x^2-(x^2-2x))`(Nhân liên hợp)

`=((x+\sqrt{x^2-2x})^2-(x-\sqrt{x^2-2x})^2)/(2x)`

`=(x^2+x^2-2x-x^2-x^2+2x+2x\sqrt{x^2-2x}+2x\sqrt{x^2-2x})/(2x)`

`=(4x\sqrt{x^2-2x})/(2x)`

`=2\sqrt{x^2-2x}`

`c)A<2`

`<=>\sqrt{x^2-2x}<1`

`<=>x^2-2x<1`

`<=>x^2-2x+1<2`

`<=>(x-1)^2<2`

`<=>-\sqrt2<x-1<\sqrt2`

`<=>1-sqrt2<x<1+sqrt2`

Kết hợp đkxđ ta có:\(\left[ \begin{array}{l}2 \le x<1+\sqrt2\\1-\sqrt2<x<0\end{array} \right.\) 

Đáp án:

a) \(\left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\)

b) \(\sqrt {{x^2} - 2x} \)

 c) \(\left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
0 > x > 1 - \sqrt 2 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
\dfrac{{x + \sqrt {{x^2} - 2x} }}{{x - \sqrt {{x^2} - 2x} }} - \dfrac{{x - \sqrt {{x^2} - 2x} }}{{x + \sqrt {{x^2} - 2x} }}\\
a)DK:\left\{ \begin{array}{l}
x - \sqrt {{x^2} - 2x}  \ne 0\\
{x^2} - 2x \ge 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
{x^2} \ne {x^2} - 2x\\
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
b)A = \dfrac{{x + \sqrt {{x^2} - 2x} }}{{x - \sqrt {{x^2} - 2x} }} - \dfrac{{x - \sqrt {{x^2} - 2x} }}{{x + \sqrt {{x^2} - 2x} }}\\
 = \dfrac{{{x^2} + 2x\sqrt {{x^2} - 2x}  + {x^2} - 2x - \left( {{x^2} - 2x\sqrt {{x^2} - 2x}  + {x^2} - 2x} \right)}}{{\left( {x - \sqrt {{x^2} - 2x} } \right)\left( {x + \sqrt {{x^2} - 2x} } \right)}}\\
 = \dfrac{{2{x^2} - 2x + 2x\sqrt {{x^2} - 2x}  - 2{x^2} + 2x + 2x\sqrt {{x^2} - 2x} }}{{\left( {x - \sqrt {{x^2} - 2x} } \right)\left( {x + \sqrt {{x^2} - 2x} } \right)}}\\
 = \dfrac{{4x\sqrt {{x^2} - 2x} }}{{{x^2} - \left( {{x^2} - 2x} \right)}}\\
 = \dfrac{{4x\sqrt {{x^2} - 2x} }}{{2x}}\\
 = 2\sqrt {{x^2} - 2x} \\
c)A < 2\\
 \to 2\sqrt {{x^2} - 2x}  < 2\\
 \to \sqrt {{x^2} - 2x}  < 1\\
 \to {x^2} - 2x < 1\\
 \to {x^2} - 2x + 1 < 2\\
 \to {\left( {x - 1} \right)^2} < 2\\
 \to \left[ \begin{array}{l}
x - 1 < \sqrt 2 \\
x - 1 >  - \sqrt 2 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x < 1 + \sqrt 2 \\
x > 1 - \sqrt 2 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
0 > x > 1 - \sqrt 2 
\end{array} \right.
\end{array}\)