Giải thích các bước giải:
$\begin{array}{l}
1b)\\
\left( {\dfrac{1}{{7 - \sqrt 5 }} - \dfrac{1}{{\sqrt 5 + 7}}} \right).\dfrac{{11\sqrt 5 - 55}}{{\sqrt 5 - 1}}\\
= \left( {\dfrac{{7 + \sqrt 5 - \left( {7 - \sqrt 5 } \right)}}{{\left( {7 - \sqrt 5 } \right)\left( {7 + \sqrt 5 } \right)}}} \right).\dfrac{{11\sqrt 5 \left( {1 - \sqrt 5 } \right)}}{{\sqrt 5 - 1}}\\
= \dfrac{{2\sqrt 5 }}{{{7^2} - {{\left( {\sqrt 5 } \right)}^2}}}.\left( { - 11\sqrt 5 } \right)\\
= \dfrac{5}{2}\\
2c)DK:x,y \ge 0\\
\dfrac{{x\sqrt x - y\sqrt y }}{{x + \sqrt {xy} + y}}\\
= \dfrac{{\left( {x\sqrt x - y\sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}{{\left( {x + \sqrt {xy} + y} \right)\left( {\sqrt x - \sqrt y } \right)}}\left( {DK:x \ne y} \right)\\
= \dfrac{{\left( {x\sqrt x - y\sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}{{{{\left( {\sqrt x } \right)}^3} - {{\left( {\sqrt y } \right)}^3}}}\\
= \dfrac{{\left( {x\sqrt x - y\sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}{{x\sqrt x - y\sqrt y }}\\
= \sqrt x - \sqrt y
\end{array}$
