Đáp án:
$\begin{array}{l}
d)\dfrac{{\sqrt {mn} }}{{\sqrt m + \sqrt n }}\\
e)\dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}\\
g)2
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
d)\dfrac{{m\sqrt n - n\sqrt m }}{{m - n}}\left( {DK:m,n > 0;m \ne n} \right)\\
= \dfrac{{\sqrt {mn} \left( {\sqrt m - \sqrt n } \right)}}{{\left( {\sqrt m - \sqrt n } \right)\left( {\sqrt m + \sqrt n } \right)}}\\
= \dfrac{{\sqrt {mn} }}{{\sqrt m + \sqrt n }}\\
e)\dfrac{{x\sqrt x - y\sqrt y }}{{x - y}}\left( {DK:x,y > 0;x \ne y} \right)\\
= \dfrac{{{{\left( {\sqrt x } \right)}^3} - {{\left( {\sqrt y } \right)}^3}}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}\\
g)\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 + 1}} - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \dfrac{{\sqrt 3 \left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}} - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \sqrt 3 + 2 + \sqrt 2 - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= 2
\end{array}$
