Đáp án:
\(\dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {\dfrac{2}{{\sqrt 3 - 1}} + \dfrac{3}{{\sqrt 3 - 2}} + \dfrac{{15}}{{3 - \sqrt 3 }}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2\sqrt 3 + 2}}{{3 - 1}} + \dfrac{{3\sqrt 3 + 6}}{{3 - 4}} + \dfrac{{45 + 15\sqrt 3 }}{{9 - 3}}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2\sqrt 3 + 2}}{2} - \dfrac{{3\sqrt 3 + 6}}{1} + \dfrac{{45 + 15\sqrt 3 }}{6}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\sqrt 3 + 1 - 3\sqrt 3 - 6 + \dfrac{{45 + 15\sqrt 3 }}{6}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left[ {\dfrac{{6\left( { - 2\sqrt 3 - 5} \right) + 45 + 15\sqrt 3 }}{6}} \right].\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{{ - 12\sqrt 3 - 30 + 45 + 15\sqrt 3 }}{6}.\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{{3\sqrt 3 + 15}}{6}.\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{{3\left( {\sqrt 3 + 5} \right)}}{6}.\dfrac{1}{{\sqrt 3 + 5}}\\
= \dfrac{3}{6} = \dfrac{1}{2}
\end{array}\)
