Đáp án:
\(d. - \dfrac{{\sqrt 5 }}{{\sqrt 3 }}\)
Giải thích các bước giải:
\(\begin{array}{l}
d.\left( {\dfrac{{\sqrt 5 - \sqrt 3 + \sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right):\left( {\dfrac{{\sqrt 5 - \sqrt 3 - \sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right)\\
= \dfrac{{2\sqrt 5 }}{{\sqrt 5 + \sqrt 3 }}.\dfrac{{\sqrt 5 + \sqrt 3 }}{{ - 2\sqrt 3 }}\\
= - \dfrac{{\sqrt 5 }}{{\sqrt 3 }}\\
f.\dfrac{{{{\left( {5\sqrt 2 - 2\sqrt 3 } \right)}^2} + {{\left( {5\sqrt 2 + 2\sqrt 3 } \right)}^2}}}{{\left( {5\sqrt 2 - 2\sqrt 3 } \right)\left( {5\sqrt 2 + 2\sqrt 3 } \right)}}\\
= \dfrac{{50 - 20\sqrt 6 + 12 + 50 + 20\sqrt 6 + 12}}{{50 - 12}}\\
= \dfrac{{124}}{{38}} = \dfrac{{62}}{{19}}\\
g.\dfrac{3}{{\sqrt 5 - \sqrt 2 }} - \dfrac{1}{{\sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} }} - \dfrac{2}{{\sqrt {8 + 2\sqrt {15} } }}\\
= \dfrac{3}{{\sqrt 5 - \sqrt 2 }} - \dfrac{1}{{\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} }} - \dfrac{2}{{\sqrt {5 + 2\sqrt 5 .\sqrt 3 + 3} }}\\
= \dfrac{3}{{\sqrt 5 - \sqrt 2 }} - \dfrac{1}{{\sqrt 3 - \sqrt 2 }} - \dfrac{2}{{\sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} }}\\
= \dfrac{3}{{\sqrt 5 - \sqrt 2 }} - \dfrac{1}{{\sqrt 3 - \sqrt 2 }} - \dfrac{2}{{\sqrt 5 + \sqrt 3 }}\\
= \dfrac{{3\sqrt 3 - 3\sqrt 2 - \sqrt 5 + \sqrt 2 }}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)}} - \dfrac{2}{{\sqrt 5 + \sqrt 3 }}\\
= \dfrac{{3\sqrt 3 - 2\sqrt 2 - \sqrt 5 }}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)}} - \dfrac{2}{{\sqrt 5 + \sqrt 3 }}\\
= \dfrac{{3\sqrt {15} - 2\sqrt {10} - 5 + 9 - 2\sqrt 6 - \sqrt {15} - 2\left( {\sqrt {15} - \sqrt {10} - \sqrt 6 + 2} \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt {15} }}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}\\
h.\left[ {\dfrac{4}{{\sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right)}} - \dfrac{{12}}{{\sqrt 3 \left( {\sqrt 3 - \sqrt 2 } \right)}} + \dfrac{{15}}{{\sqrt 6 + 1}}} \right].\left( {\sqrt 6 + 11} \right)\\
= \left[ {\dfrac{{2\sqrt 2 - 4\sqrt 3 }}{{\sqrt 3 - \sqrt 2 }} + \dfrac{{15}}{{\sqrt 6 + 1}}} \right].\left( {\sqrt 6 + 11} \right)\\
= \dfrac{{2.2\sqrt 3 - 12\sqrt 2 + 15\sqrt 3 - 15\sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 + 1} \right)}}.\left( {\sqrt 6 + 11} \right)\\
= \dfrac{{19\sqrt 3 - 27\sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 + 1} \right)}}.\left( {\sqrt 6 + 11} \right)\\
= \dfrac{{57\sqrt 2 - 54\sqrt 3 + 209\sqrt 3 - 297\sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 + 1} \right)}}\\
= \dfrac{{ - 240\sqrt 2 + 155\sqrt 3 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 + 1} \right)}}
\end{array}\)
