Đáp án:

\(\begin{array}{l}
1,\\
B = \dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
2,\\
A = \dfrac{1}{{\sqrt x  - 3}}\\
3,\\
P = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
4,\\
C = \dfrac{1}{{\sqrt a  + 1}}
\end{array}\) 

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,x > 0\\
B = \dfrac{{\sqrt x  - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x  + 1}}{{x + \sqrt x }}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x  + 1}}{{{{\sqrt x }^2} + \sqrt x }}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right) + 2\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} - {1^2} + 2\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} + 2\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
2,\\
A = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{3}{{\sqrt x  - 3}}} \right).\dfrac{{\sqrt x  + 3}}{{x + 9}}\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  - 3} \right) + 3.\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  + 3}}{{x + 9}}\\
 = \dfrac{{x - 3\sqrt x  + 3\sqrt x  + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  + 3}}{{x + 9}}\\
 = \dfrac{{x + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  + 3}}{{x + 9}}\\
 = \dfrac{1}{{\sqrt x  - 3}}\\
3,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
P = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} - \dfrac{1}{{x - \sqrt x }}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} - \dfrac{1}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
4,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right.\\
C = \dfrac{{a - \sqrt a }}{{a - 1}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}\\
 = \dfrac{{{{\sqrt a }^2} - \sqrt a }}{{{{\sqrt a }^2} - {1^2}}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a .\left( {\sqrt a  - 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a }}{{\sqrt a  + 1}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a  - \left( {\sqrt a  - 1} \right)}}{{\sqrt a  + 1}}\\
 = \dfrac{1}{{\sqrt a  + 1}}
\end{array}\)