Đáp án:

 \(\begin{array}{l}
A = 4\\
B = 6x\\
C = \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}}\\
D = \dfrac{1}{{\sqrt a }}\\
E = \dfrac{{1 - x}}{{\sqrt x }}\\
F = \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}}\\
G = \dfrac{{x + 4}}{{\sqrt x }}
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
*)\\
A = \left( {\dfrac{1}{{\sqrt x  - 2}} - \dfrac{1}{{\sqrt x  + 2}} + \dfrac{{4\sqrt x }}{{x - 4}}} \right):\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \left( {\dfrac{1}{{\sqrt x  - 2}} - \dfrac{1}{{\sqrt x  + 2}} + \dfrac{{4\sqrt x }}{{{{\sqrt x }^2} - {2^2}}}} \right):\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \left( {\dfrac{1}{{\sqrt x  - 2}} - \dfrac{1}{{\sqrt x  + 2}} + \dfrac{{4\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}} \right):\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \dfrac{{\left( {\sqrt x  + 2} \right) - \left( {\sqrt x  - 2} \right) + 4\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \dfrac{{\sqrt x  + 2 - \sqrt x  + 2 + 4\sqrt x }}{{{{\sqrt x }^2} - {2^2}}}:\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \dfrac{{4\sqrt x  + 4}}{{x - 4}}:\dfrac{{\sqrt x  + 1}}{{x - 4}}\\
 = \dfrac{{4.\left( {\sqrt x  + 1} \right)}}{{x - 4}}.\dfrac{{x - 4}}{{\sqrt x  + 1}}\\
 = 4\\
*)\\
B = \left( {\dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} + \dfrac{{6x}}{{x - 1}}} \right).\dfrac{{x\sqrt x  - \sqrt x }}{{\sqrt x  - 1}}\\
 = \left( {\dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} + \dfrac{{6x}}{{{{\sqrt x }^2} - {1^2}}}} \right).\dfrac{{x\sqrt x  - \sqrt x }}{{\sqrt x  - 1}}\\
 = \left( {\dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} + \dfrac{{6x}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}} \right).\dfrac{{x\sqrt x  - \sqrt x }}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 1} \right) + 6x}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x .\left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {x - 3\sqrt x  + 2} \right) - \left( {x + 3\sqrt x  + 2} \right) + 6x}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x \left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{x - 3\sqrt x  + 2 - x - 3\sqrt x  - 2 + 6x}}{{x - 1}}.\dfrac{{\sqrt x \left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{6x - 6\sqrt x }}{{x - 1}}.\dfrac{{\sqrt x \left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{6\sqrt x \left( {\sqrt x  - 1} \right)}}{{x - 1}}.\dfrac{{\sqrt x .\left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = 6\sqrt x .\sqrt x \\
 = 6x\\
*)\\
C = \left( {\dfrac{{x\sqrt x  + x + \sqrt x }}{{x\sqrt x  - 1}} - \dfrac{{\sqrt x  + 3}}{{1 - \sqrt x }}} \right).\dfrac{{x - 1}}{{2x + \sqrt x  - 1}}\\
 = \left( {\dfrac{{\sqrt x .\left( {x + \sqrt x  + 1} \right)}}{{{{\sqrt x }^3} - {1^3}}} - \dfrac{{\sqrt x  + 3}}{{1 - \sqrt x }}} \right).\dfrac{{{{\sqrt x }^2} - {1^2}}}{{\left( {2x + 2\sqrt x } \right) + \left( { - \sqrt x  - 1} \right)}}\\
 = \left( {\dfrac{{\sqrt x \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 1}}} \right).\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{2\sqrt x \left( {\sqrt x  + 1} \right) - \left( {\sqrt x  + 1} \right)}}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 1}}} \right).\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {2\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  + \sqrt x  + 3}}{{\sqrt x  - 1}}.\dfrac{{\sqrt x  - 1}}{{2\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 3}}{{\sqrt x  - 1}}.\dfrac{{\sqrt x  - 1}}{{2\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}}\\
*)\\
D = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{a\sqrt a  - a - \sqrt a  + 1}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{\left( {a\sqrt a  - a} \right) - \left( {\sqrt a  - 1} \right)}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{a\left( {\sqrt a  - 1} \right) - \left( {\sqrt a  - 1} \right)}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \left( {\dfrac{6}{{a - 1}} + \dfrac{{10 - 2\sqrt a }}{{\left( {\sqrt a  - 1} \right).\left( {a - 1} \right)}}} \right).\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{6.\left( {\sqrt a  - 1} \right) + \left( {10 - 2\sqrt a } \right)}}{{\left( {\sqrt a  - 1} \right).\left( {a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{6\sqrt a  - 6 + 10 - 2\sqrt a }}{{\left( {\sqrt a  - 1} \right).\left( {\sqrt a  - 1} \right).\left( {\sqrt a  + 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{4\sqrt a  + 4}}{{{{\left( {\sqrt a  - 1} \right)}^2}.\left( {\sqrt a  + 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{{4.\left( {\sqrt a  + 1} \right)}}{{{{\left( {\sqrt a  - 1} \right)}^2}.\left( {\sqrt a  + 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{4\sqrt a }}\\
 = \dfrac{1}{{\sqrt a }}\\
*)\\
E = \left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}} \right).{\left( {\dfrac{1}{{2\sqrt x }} - \dfrac{{\sqrt x }}{2}} \right)^2}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2} - {{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  + 1} \right).\left( {\sqrt x  - 1} \right)}}.{\left( {\dfrac{{1 - \sqrt x .\sqrt x }}{{2\sqrt x }}} \right)^2}\\
 = \dfrac{{\left( {x - 2\sqrt x  + 1} \right) - \left( {x + 2\sqrt x  + 1} \right)}}{{{{\sqrt x }^2} - {1^2}}}.{\left( {\dfrac{{1 - x}}{{2\sqrt x }}} \right)^2}\\
 = \dfrac{{x - 2\sqrt x  + 1 - x - 2\sqrt x  - 1}}{{x - 1}}.\dfrac{{{{\left( {1 - x} \right)}^2}}}{{{{\left( {2\sqrt x } \right)}^2}}}\\
 = \dfrac{{ - 4\sqrt x }}{{x - 1}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{4x}}\\
 = \dfrac{{ - \left( {x - 1} \right)}}{{\sqrt x }}\\
 = \dfrac{{1 - x}}{{\sqrt x }}\\
*)\\
F = \left( {\dfrac{{3\sqrt x  + 5}}{{x - \sqrt x  - 2}} + \dfrac{{2\sqrt x  + 4}}{{4 - x}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
 = \left( {\dfrac{{3\sqrt x  + 5}}{{\left( {x - 2\sqrt x } \right) + \left( {\sqrt x  - 2} \right)}} - \dfrac{{2\sqrt x  + 4}}{{x - 4}}} \right):\dfrac{{\sqrt x }}{{{{\sqrt x }^2} + \sqrt x }}\\
 = \left( {\dfrac{{3\sqrt x  + 5}}{{\sqrt x \left( {\sqrt x  - 2} \right) + \left( {\sqrt x  - 2} \right)}} - \dfrac{{2\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}} \right):\dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \left( {\dfrac{{3\sqrt x  + 5}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}} - \dfrac{2}{{\sqrt x  - 2}}} \right):\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {3\sqrt x  + 5} \right) - 2.\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{3\sqrt x  + 5 - 2\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}}\\
*)\\
G = \left( {\dfrac{{x\sqrt x  - x}}{{x - 1}} + \dfrac{{4\sqrt x }}{{x + \sqrt x }}} \right):\dfrac{{\sqrt x }}{{\sqrt x  + 1}}\\
 = \left( {\dfrac{{x\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} + \dfrac{{4\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} \right).\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \left( {\dfrac{x}{{\sqrt x  + 1}} + \dfrac{4}{{\sqrt x  + 1}}} \right).\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{x + 4}}{{\sqrt x  + 1}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = \dfrac{{x + 4}}{{\sqrt x }}
\end{array}\)