Đáp án:
$\begin{array}{l}
a)4 - \sqrt {15} \\
b)\dfrac{{a - 2\sqrt a + 1}}{{a - 1}},a \ge 0;a \ne 1
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)\dfrac{{5\sqrt 3 - 3\sqrt 5 }}{{5\sqrt 3 + 3\sqrt 5 }}\\
= \dfrac{{\left( {5\sqrt 3 - 3\sqrt 5 } \right)\left( {5\sqrt 3 - 3\sqrt 5 } \right)}}{{\left( {5\sqrt 3 + 3\sqrt 5 } \right)\left( {5\sqrt 3 - 3\sqrt 5 } \right)}}\\
= \dfrac{{{{\left( {5\sqrt 3 } \right)}^2} - 2.5\sqrt 3 .3\sqrt 5 + {{\left( {3\sqrt 5 } \right)}^2}}}{{{{\left( {5\sqrt 3 } \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}}}\\
= \dfrac{{120 - 30\sqrt {15} }}{{30}}\\
= 4 - \sqrt {15} \\
b)DK:a \ge 0;a \ne 1\\
\dfrac{{1 - \sqrt a }}{{1 + \sqrt a }}\\
= \dfrac{{\left( {1 - \sqrt a } \right)\left( {1 - \sqrt a } \right)}}{{\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)}}\\
= \dfrac{{a - 2\sqrt a + 1}}{{a - 1}}
\end{array}$
