Đáp án:
\(\begin{array}{l}
a) - 11\\
b) - \dfrac{{23\sqrt {10} }}{{15}}\\
c)\dfrac{1}{{\sqrt 5 + 2}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}} - \sqrt 6 \\
= 3\left( {\sqrt 6 - 1} \right) + 2\left( {\sqrt 6 + 2} \right) - 4\left( {3 + \sqrt 6 } \right) - \sqrt 6 \\
= 3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 - \sqrt 6 \\
= - 11\\
b)\dfrac{{{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2} - {{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}}}{{5 - 2}} - \dfrac{2}{{\sqrt {10} }}\\
= \dfrac{{7 - 2\sqrt {10} - 7 - 2\sqrt {10} }}{3} - \dfrac{2}{{\sqrt {10} }}\\
= \dfrac{{ - 4\sqrt {10} }}{3} - \dfrac{2}{{\sqrt {10} }} = - \dfrac{{23\sqrt {10} }}{{15}}\\
c)\dfrac{{\sqrt 5 - 2 - \sqrt 5 + \sqrt 5 + 2}}{{\sqrt 5 \left( {\sqrt 5 + 2} \right)}}\\
= \dfrac{{\sqrt 5 }}{{\sqrt 5 \left( {\sqrt 5 + 2} \right)}} = \dfrac{1}{{\sqrt 5 + 2}}
\end{array}\)
