Đáp án:
c. \(\dfrac{1}{{\sqrt 5 }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{\sqrt 5 - 1 + \sqrt 5 + 1}}{{5 - 1}} - \dfrac{{3 + \sqrt 5 }}{{9 - 5}} - \sqrt 5 \\
= \dfrac{{2\sqrt 5 }}{4} - \dfrac{{3 + \sqrt 5 }}{4} - \sqrt 5 \\
= \dfrac{{2\sqrt 5 - 3 - \sqrt 5 - 4\sqrt 5 }}{4}\\
= \dfrac{{ - 3 - 3\sqrt 5 }}{4}\\
b.\left( {\dfrac{{15\sqrt 6 - 15}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}}} \right)\left( {\sqrt 6 + 11} \right)\\
= \left[ {\dfrac{{15\sqrt 6 - 15}}{5} + \dfrac{{4\sqrt 6 + 8}}{2} - \dfrac{{36 + 12\sqrt 6 }}{3}} \right]\left( {\sqrt 6 + 11} \right)\\
= \left( {3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 } \right)\left( {\sqrt 6 + 11} \right)\\
= \left( {\sqrt 6 - 11} \right)\left( {\sqrt 6 + 11} \right)\\
= 6 - 121 = - 115\\
c.2.\dfrac{4}{{\sqrt 5 }} - \dfrac{3}{{3\sqrt 5 }} - \dfrac{6}{{\sqrt 5 }} = \dfrac{8}{{\sqrt 5 }} - \dfrac{1}{{\sqrt 5 }} - \dfrac{6}{{\sqrt 5 }}\\
= \dfrac{1}{{\sqrt 5 }}
\end{array}\)
