Đáp án:
$A = \dfrac{{15\sqrt {10} - 5\sqrt 2 }}{{22}}$
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{{3 + \sqrt 5 }}{{\sqrt {10} + \sqrt {3 + \sqrt 5 } }} - \dfrac{{3 - \sqrt 5 }}{{\sqrt {10} - \sqrt {3 - \sqrt 5 } }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2\sqrt 5 + \sqrt {6 + 2\sqrt 5 } }} - \dfrac{{\sqrt 2 \left( {3 - \sqrt 5 } \right)}}{{2\sqrt 5 - \sqrt {6 - 2\sqrt 5 } }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2\sqrt 5 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} - \dfrac{{\sqrt 2 \left( {3 - \sqrt 5 } \right)}}{{2\sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt 5 + 1}} - \dfrac{{3\sqrt 2 - \sqrt {10} }}{{2\sqrt 5 - \left( {\sqrt 5 - 1} \right)}}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{3\sqrt 5 + 1}} - \dfrac{{3\sqrt 2 - \sqrt {10} }}{{1 - \sqrt 5 }}\\
= \dfrac{{\left( {3\sqrt 2 + \sqrt {10} } \right)\left( {3\sqrt 5 - 1} \right)}}{{\left( {3\sqrt 5 + 1} \right)\left( {3\sqrt 5 - 1} \right)}} - \dfrac{{\left( {3\sqrt 2 - \sqrt {10} } \right)\left( {1 + \sqrt 5 } \right)}}{{\left( {1 - \sqrt 5 } \right)\left( {1 + \sqrt 5 } \right)}}\\
= \dfrac{{9\sqrt {10} - 3\sqrt 2 + 15\sqrt 2 - \sqrt {10} }}{{{{\left( {3\sqrt 5 } \right)}^2} - {1^2}}} - \dfrac{{3\sqrt 2 + 3\sqrt {10} - \sqrt {10} - 5\sqrt 2 }}{{{1^2} - {{\left( {\sqrt 5 } \right)}^2}}}\\
= \dfrac{{12\sqrt 2 + 8\sqrt {10} }}{{44}} - \dfrac{{2\sqrt {10} - 2\sqrt 2 }}{{ - 4}}\\
= \dfrac{{3\sqrt 2 + 2\sqrt {10} }}{{11}} + \dfrac{{\sqrt {10} - \sqrt 2 }}{2}\\
= \dfrac{{2\left( {3\sqrt 2 + 2\sqrt {10} } \right) + 11\left( {\sqrt {10} - \sqrt 2 } \right)}}{{22}}\\
= \dfrac{{6\sqrt 2 + 4\sqrt {10} + 11\sqrt {10} - 11\sqrt 2 }}{{22}}\\
= \dfrac{{15\sqrt {10} - 5\sqrt 2 }}{{22}}
\end{array}$
Vậy $A = \dfrac{{15\sqrt {10} - 5\sqrt 2 }}{{22}}$
