Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
n,\\
N = \left( {\dfrac{{2x - 4\sqrt x  + 2}}{{x - 4}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}} \right):\dfrac{{\sqrt x }}{{\sqrt x  - 2}}\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right)\\
 = \left( {\dfrac{{2x - 4\sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} - \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 2}}} \right):\dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
 = \dfrac{{\left( {2x - 4\sqrt x  + 2} \right) - \left( {2\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
 = \dfrac{{\left( {2x - 4\sqrt x  + 2} \right) - \left( {2x - 5\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
 = \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
 = \dfrac{1}{{\sqrt x  + 2}}\\
U = \left( {\dfrac{{x - y}}{{\sqrt x  - \sqrt y }} - \dfrac{{x\sqrt x  - y\sqrt y }}{{x - y}}} \right).\dfrac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{x\sqrt x  + y\sqrt y }}\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x,y \ge 0\\
x \ne y
\end{array} \right)\\
 = \left( {\dfrac{{x - y}}{{\sqrt x  - \sqrt y }} - \dfrac{{x\sqrt x  - y\sqrt y }}{{\left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right)}}} \right).\dfrac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{x\sqrt x  + y\sqrt y }}\\
 = \dfrac{{\left( {x - y} \right)\left( {\sqrt x  + \sqrt y } \right) - \left( {x\sqrt x  - y\sqrt y } \right)}}{{\left( {\sqrt x  - \sqrt y } \right).\left( {\sqrt x  + \sqrt y } \right)}}.\dfrac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{\left( {\sqrt x  + \sqrt y } \right)\left( {x - \sqrt {xy}  + y} \right)}}\\
 = \dfrac{{\left( {x\sqrt x  + x\sqrt y  - y\sqrt x  - y\sqrt y } \right) - \left( {x\sqrt x  - y\sqrt y } \right)}}{{\left( {\sqrt x  + \sqrt y } \right)}}.\dfrac{{\left( {\sqrt x  - \sqrt y } \right)}}{{\left( {\sqrt x  + \sqrt y } \right).\left( {x - \sqrt {xy}  + y} \right)}}\\
 = \dfrac{{x\sqrt y  - y\sqrt x }}{{\sqrt x  + \sqrt y }}.\dfrac{{\left( {\sqrt x  - \sqrt y } \right)}}{{\left( {\sqrt x  + \sqrt y } \right).\left( {x - \sqrt {xy}  + y} \right)}}\\
 = \dfrac{{\sqrt {xy} .{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{{{\left( {\sqrt x  + \sqrt y } \right)}^2}.\left( {x - \sqrt {xy}  + y} \right)}}\\
V = \left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right) + \dfrac{{1 - \sqrt a }}{{1 - a}}\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right)\\
 = \left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a  + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right) + \dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}\\
 = \left( {1 + \sqrt a  + a + \sqrt a } \right) + \dfrac{1}{{1 + \sqrt a }}\\
 = \left( {a + 2\sqrt a  + 1} \right) + \dfrac{1}{{\sqrt a  + 1}}\\
 = {\left( {\sqrt a  + 1} \right)^2} + \dfrac{1}{{\sqrt a  + 1}}\\
 = \dfrac{{{{\left( {\sqrt a  + 1} \right)}^3} + 1}}{{\sqrt a  + 1}}\\
X = \dfrac{{15\sqrt x  - 11}}{{x + 2\sqrt x  - 3}} + \dfrac{{3\sqrt x  - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x  + 3}}{{3 + \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right)\\
 = \dfrac{{15\sqrt x  - 11}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}} - \dfrac{{3\sqrt x  - 2}}{{\sqrt x  - 1}} - \dfrac{{2\sqrt x  + 3}}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {15\sqrt x  - 11} \right) - \left( {3\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right) - \left( {2\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{15\sqrt x  - 11 - \left( {3x + 7\sqrt x  - 6} \right) - \left( {2x + \sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{15\sqrt x  - 11 - 3x - 7\sqrt x  + 6 - 2x - \sqrt x  + 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{ - 5x + 7\sqrt x  - 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{ - \left( {5x - 7\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{ - \left( {\sqrt x  - 1} \right)\left( {5\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2 - 5\sqrt x }}{{\sqrt x  + 3}}
\end{array}\)