Đáp án:

\(\begin{array}{l}
1,\\
\dfrac{{2\sqrt x }}{{x - 1}}\\
2,\\
\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
3,\\
\dfrac{{\sqrt x  - 1}}{{x - 4}}\\
4,\\
\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
5,\\
\dfrac{{3\sqrt x  + 3}}{{x - 9}}\\
6,\\
\dfrac{3}{{\sqrt x  - 3}}\\
7,\\
\dfrac{1}{{x + \sqrt x  + 1}}\\
8,\\
\dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x }}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\dfrac{1}{{\sqrt x  + 1}} + \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right) + \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1 + \sqrt x  + 1}}{{{{\sqrt x }^2} - {1^2}}}\\
 = \dfrac{{2\sqrt x }}{{x - 1}}\\
2,\\
\dfrac{{\sqrt x }}{{\sqrt x  - 1}} - \dfrac{1}{{x - \sqrt x }}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} - \dfrac{1}{{{{\sqrt x }^2} - \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 }}\\
3,\\
\dfrac{1}{{\sqrt x  + 2}} + \dfrac{1}{{x - 4}}\\
 = \dfrac{1}{{\sqrt x  + 2}} + \dfrac{1}{{{{\sqrt x }^2} - {2^2}}}\\
 = \dfrac{1}{{\sqrt x  + 2}} + \dfrac{1}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right) + 1}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{{{\sqrt x }^2} - {2^2}}}\\
 = \dfrac{{\sqrt x  - 1}}{{x - 4}}\\
4,\\
\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 }}\\
5,\\
\dfrac{2}{{\sqrt x  - 3}} + \dfrac{1}{{\sqrt x  + 3}}\\
 = \dfrac{{2.\left( {\sqrt x  + 3} \right) + 1.\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2\sqrt x  + 6 + \sqrt x  - 3}}{{{{\sqrt x }^2} - {3^2}}}\\
 = \dfrac{{3\sqrt x  + 3}}{{x - 9}}\\
6,\\
\dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x - 9}}{{x - 9}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x - 9}}{{{{\sqrt x }^2} - {3^2}}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  - 3} \right) + 2\sqrt x .\left( {\sqrt x  + 3} \right) - \left( {3x - 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\left( {x - 3\sqrt x } \right) + \left( {2x + 6\sqrt x } \right) - \left( {3x - 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x  + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3.\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{3}{{\sqrt x  - 3}}\\
7,\\
\dfrac{{2\sqrt x  + x}}{{x\sqrt x  - 1}} - \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + x}}{{{{\sqrt x }^3} - {1^3}}} - \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + x}}{{\left( {\sqrt x  - 1} \right)\left( {{{\sqrt x }^2} + \sqrt x .1 + {1^2}} \right)}} - \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + x}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} - \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {2\sqrt x  + x} \right) - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{2\sqrt x  + x - x - \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{1}{{x + \sqrt x  + 1}}\\
8,\\
\dfrac{{x\sqrt x  - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x  + 1}}{{x + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
 = \dfrac{{{{\sqrt x }^3} - {1^3}}}{{{{\sqrt x }^2} - \sqrt x }} - \dfrac{{{{\sqrt x }^3} + {1^3}}}{{{{\sqrt x }^2} + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {{{\sqrt x }^2} + \sqrt x .1 + {1^2}} \right)}}{{\sqrt x .\left( {\sqrt x  - 1} \right)}} - \dfrac{{\left( {\sqrt x  + 1} \right)\left( {{{\sqrt x }^2} - \sqrt x .1 + {1^2}} \right)}}{{\sqrt x .\left( {\sqrt x  + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{\sqrt x .\left( {\sqrt x  - 1} \right)}} - \dfrac{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x  + 1}}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
 = \dfrac{{\left( {x + \sqrt x  + 1} \right) - \left( {x - \sqrt x  + 1} \right) + x + 1}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  + 1 - x + \sqrt x  - 1 + x + 1}}{{\sqrt x }}\\
 = \dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x }}
\end{array}\)