Đáp án:

$\begin{array}{l}
g)\dfrac{{x - y + 3\sqrt x  + 3\sqrt y }}{{\sqrt x  - \sqrt y  + 3}}\\
 = \dfrac{{\left( {\sqrt x  + \sqrt y } \right)\left( {\sqrt x  - \sqrt y } \right) + 3\left( {\sqrt x  + \sqrt y } \right)}}{{\sqrt x  - \sqrt y  + 3}}\\
 = \dfrac{{\left( {\sqrt x  - \sqrt y  + 3} \right)\left( {\sqrt x  - \sqrt y } \right)}}{{\sqrt x  - \sqrt y  + 3}}\\
 = \sqrt x  - \sqrt y \\
i)\sqrt {x + 2\sqrt {x - 1} }  - \sqrt {x - 1} \\
 = \sqrt {x - 1 + 2\sqrt {x - 1}  + 1}  - \sqrt {x - 1} \\
 = \sqrt {{{\left( {\sqrt {x - 1}  + 1} \right)}^2}}  - \sqrt {x - 1} \\
 = \sqrt {x - 1}  + 1 - \sqrt {x - 1} \\
 = 1\\
k)\dfrac{{\sqrt {x - 2\sqrt {x - 1} }  + \sqrt {x + 2\sqrt {x - 1} } }}{{\sqrt {\dfrac{1}{{{x^2}}} - \dfrac{2}{x} + 1} }}\\
 = \dfrac{{\sqrt {{{\left( {\sqrt {x - 1}  - 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt {x - 1}  + 1} \right)}^2}} }}{{\sqrt {\dfrac{{1 - 2x + {x^2}}}{{{x^2}}}} }}\\
 = \dfrac{{\left| {\sqrt {x - 1}  - 1} \right| + \sqrt {x - 1}  + 1}}{{\left| {x - 1} \right|}}.\left| x \right|\\
 = \left[ \begin{array}{l}
\dfrac{{\sqrt {x - 1}  - 1 + \sqrt {x - 1}  + 1}}{{x - 1}}.x = \dfrac{{2x}}{{\sqrt {x - 1} }}\left( {khi:x \ge 2} \right)\\
\dfrac{{1 - \sqrt {x - 1}  + \sqrt {x - 1}  + 1}}{{x - 1}}.x = \dfrac{2}{{x - 1}}\left( {khi:1 < x < 2} \right)
\end{array} \right.\\
m)\\
\left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \dfrac{{\sqrt x .\left( {3 - \sqrt x } \right) + x + 9}}{{\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}:\dfrac{{3\sqrt x  + 1 - \left( {\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  - x + x + 9}}{{\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{3\left( {\sqrt x  + 3} \right)}}{{3 + \sqrt x }}.\dfrac{{ - \sqrt x }}{{2\sqrt x  + 4}}\\
 = \dfrac{{ - 3\sqrt x }}{{2\sqrt x  + 4}}\\
h)\sqrt {x - \sqrt {{x^2} - 4} } .\sqrt {x + \sqrt {{x^2} - 4} } \\
 = \sqrt {{x^2} - \left( {{x^2} - 4} \right)} \\
 = \sqrt 4 \\
 = 2\\
j)\\
\left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}}  - 1}} - \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}}} \right).\left( {\dfrac{{1 + \sqrt {{x^3}} }}{{1 + \sqrt x }} - \sqrt x } \right)\\
 = \left( {\dfrac{{2x + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} - \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}}} \right)\\
.\left( {\dfrac{{\left( {1 + \sqrt x } \right)\left( {1 - \sqrt x  + x} \right)}}{{1 + \sqrt x }} - \sqrt x } \right)\\
 = \dfrac{{2x + 1 - \sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\left( {x - 2\sqrt x  + 1} \right)\\
 = \dfrac{{2x + 1 - x + \sqrt x }}{{x + \sqrt x  + 1}}.{\left( {\sqrt x  - 1} \right)^2}\\
 = \dfrac{{x + \sqrt x  + 1}}{{x + \sqrt x  + 1}}.{\left( {\sqrt x  - 1} \right)^2}\\
 = {\left( {\sqrt x  - 1} \right)^2}\\
l)\left( {\dfrac{{x\sqrt x  + y\sqrt y }}{{\sqrt x  + \sqrt y }} - 2\sqrt {xy} } \right):\left( {x - y} \right) + \dfrac{{2\sqrt y }}{{\sqrt x  + \sqrt y }}\\
 = \left( {x - \sqrt {xy}  + y - 2\sqrt {xy} } \right):\left( {x - y} \right) + \dfrac{{2\sqrt y }}{{\sqrt x  + \sqrt y }}\\
 = \dfrac{{x - 3\sqrt {xy}  + y}}{{\left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right)}} + \dfrac{{2\sqrt y }}{{\sqrt x  + \sqrt y }}\\
 = \dfrac{{x - 3\sqrt {xy}  + y + 2\sqrt y \left( {\sqrt x  - \sqrt y } \right)}}{{\left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right)}}\\
 = \dfrac{{x - \sqrt {xy}  - y}}{{x - y}}
\end{array}$