Đáp án:

$\begin{array}{l}
1)b)Dkxd:x >  - 2\\
\dfrac{{3x + 2}}{{\sqrt {x + 2} }} = 2\sqrt {x + 2} \\
 \Rightarrow 3x + 2 = 2\left( {x + 2} \right)\\
 \Rightarrow 3x + 2 = 2x + 4\\
 \Rightarrow x = 2\left( {tmdk} \right)\\
Vay\,x = 2\\
2)Dkxd:a > 0;b > 0;a \ne b\\
a)A = \dfrac{{a + b - \sqrt {ab} }}{{a\sqrt a  + b\sqrt b }} - \dfrac{{\sqrt a  - \sqrt b  - 1}}{{a - b}}\\
 = \dfrac{{a - \sqrt {ab}  + b}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {a - \sqrt {ab}  + b} \right)}} - \dfrac{{\sqrt a  - \sqrt b  - 1}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  - \sqrt b } \right)}}\\
 = \dfrac{1}{{\sqrt a  + \sqrt b }} - \dfrac{{\sqrt a  - \sqrt b  - 1}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  - \sqrt b } \right)}}\\
 = \dfrac{{\sqrt a  - \sqrt b  - \sqrt a  + \sqrt b  + 1}}{{\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  - \sqrt b } \right)}}\\
 = \dfrac{1}{{a - b}}\\
b)a - b = 1\\
 \Rightarrow A = \dfrac{1}{1} = 1\\
28)a)Dkxd:2{x^2} - 3x + 1 \ge 0\\
 \Rightarrow \left( {2x - 1} \right)\left( {x - 1} \right) \ge 0\\
 \Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow A\\
b)Dkxd:\left\{ \begin{array}{l}
x - 1 \ge 0\\
2x - 1 \ge 0
\end{array} \right. \Rightarrow x \ge 1\\
 \Rightarrow B\\
c)A = B\\
 \Rightarrow \sqrt {2{x^2} - 3x + 1}  = \sqrt {x - 1} .\sqrt {2x - 1} \\
 \Rightarrow \sqrt {\left( {x - 1} \right)\left( {2x - 1} \right)}  = \sqrt {x - 1} .\sqrt {2x - 1} \\
 \Rightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
2x - 1 \ge 0
\end{array} \right. \Rightarrow x \ge 1\\
 \Rightarrow C\\
d)x \le \dfrac{1}{2} \Rightarrow D
\end{array}$