Giải thích các bước giải:
$\begin{array}{l}
a)4 - \left( {\dfrac{1}{2}\sqrt {28} + 2\sqrt {\dfrac{7}{9}} - \sqrt {175} } \right):\sqrt 7 \\
= 4 - \left( {\dfrac{1}{2}.2\sqrt 7 + 2.\dfrac{{\sqrt 7 }}{3} - 5\sqrt 7 } \right):\sqrt 7 \\
= 4 - \left( {1 + \dfrac{2}{3} - 5} \right)\\
= \dfrac{{22}}{3}\\
b)\dfrac{{\sqrt 5 + 1}}{{\sqrt 5 + 2}} - \dfrac{{\sqrt 5 - 1}}{{\sqrt 5 - 2}}\\
= \dfrac{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 - 2} \right) - \left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 2} \right)}}{{\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)}}\\
= \dfrac{{3 - \sqrt 5 - \left( {3 + \sqrt 5 } \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {2^2}}}\\
= \dfrac{{ - 2\sqrt 5 }}{{21}}\\
c)\dfrac{1}{{7 + 4\sqrt 3 }} + \dfrac{1}{{7 - 4\sqrt 3 }}\\
= \dfrac{{7 - 4\sqrt 3 + 7 + 4\sqrt 3 }}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}\\
= \dfrac{{14}}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}}\\
= 14\\
d)\left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\left( {DK:x \ge 0,x \ne 9} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}
\end{array}$
