Đáp án:
\(\begin{array}{l}
a) - \dfrac{{18\sqrt 3 }}{{19}}\\
b)3 - \sqrt {10} \\
c)\dfrac{1}{2}\\
d)14\\
e)\dfrac{{15 - 3\sqrt 3 }}{{22}}\\
n)0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{3\left( {2\sqrt 2 + 3\sqrt 3 } \right) - 3\left( {2\sqrt 2 - 3\sqrt 3 } \right)}}{{8 - 27}}\\
= \dfrac{{6\sqrt 2 + 9\sqrt 3 - 6\sqrt 2 + 9\sqrt 3 }}{{ - 19}}\\
= - \dfrac{{18\sqrt 3 }}{{19}}\\
b)\dfrac{{{{\left( {5 + \sqrt 5 } \right)}^2} + {{\left( {5 - \sqrt 5 } \right)}^2}}}{{25 - 5}} - \sqrt {10} \\
= \dfrac{{25 + 10\sqrt 5 + 5 + 25 - 10\sqrt 5 + 5}}{{20}} - \sqrt {10} \\
= \dfrac{{60}}{{20}} - \sqrt {10} = 3 - \sqrt {10} \\
c)\dfrac{{3 + \sqrt 5 - 3 + \sqrt 5 }}{{9 - 5}}.\dfrac{{\sqrt 5 - 1}}{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}\\
= \dfrac{{2\sqrt 5 }}{{4.\sqrt 5 }} = \dfrac{1}{2}\\
d)\dfrac{{7 - 4\sqrt 3 + 7 + 4\sqrt 3 }}{{49 - 48}}\\
= 14\\
e)\left[ {\dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{3 - 4}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{{9 - 3}}} \right].\dfrac{1}{{5 + \sqrt 3 }}\\
= \left( {\sqrt 3 + 1 - 3\left( {\sqrt 3 + 2} \right) + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{6}} \right).\dfrac{1}{{5 + \sqrt 3 }}\\
= \left( { - 2\sqrt 3 - 1 + \dfrac{5}{3}\left( {3 + \sqrt 3 } \right)} \right).\dfrac{1}{{5 + \sqrt 3 }}\\
= \left( { - \dfrac{{\sqrt 3 }}{3} + 4} \right).\dfrac{1}{{5 + \sqrt 3 }}\\
= \dfrac{{ - \sqrt 3 + 12}}{{15 + 3\sqrt 3 }} = \dfrac{{15 - 3\sqrt 3 }}{{22}}\\
n)\dfrac{1}{{\sqrt {7 - 2\sqrt 6 } + 1}} - \dfrac{1}{{\sqrt {7 + 2\sqrt 6 } - 1}}\\
= \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2} + 1} }} - \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 + 1} \right)}^2} - 1} }}\\
= \dfrac{1}{{\sqrt 6 - 1 + 1}} - \dfrac{1}{{\sqrt 6 + 1 - 1}}\\
= \dfrac{1}{{\sqrt 6 }} - \dfrac{1}{{\sqrt 6 }} = 0
\end{array}\)
