Đáp án:
a. -115
Giải thích các bước giải:
\(\begin{array}{l}
C15:\\
a.\left[ {\dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}}} \right].\left( {\sqrt 6 + 11} \right)\\
= \left[ {\dfrac{{15\left( {\sqrt 6 - 1} \right)}}{5} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{2} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{3}} \right].\left( {\sqrt 6 + 11} \right)\\
= \left[ {3\left( {\sqrt 6 - 1} \right) + 2\left( {\sqrt 6 + 2} \right) - 4\left( {3 + \sqrt 6 } \right)} \right].\left( {\sqrt 6 + 11} \right)\\
= \left( {3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 } \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {\sqrt 6 - 11} \right).\left( {\sqrt 6 + 11} \right)\\
= 6 - {11^2} = - 115\\
c.\dfrac{{\sqrt 5 - 1}}{{5 - 1}} + \dfrac{{\sqrt 5 + 2}}{{5 - 4}} - \dfrac{{3 + \sqrt 5 }}{{9 - 5}} - \sqrt 5 \\
= \dfrac{{\sqrt 5 - 1}}{4} + \dfrac{{\sqrt 5 + 2}}{1} - \dfrac{{3 + \sqrt 5 }}{4} - \sqrt 5 \\
= \dfrac{{\sqrt 5 - 1 + 4\sqrt 5 + 8 - 3 - \sqrt 5 - 4\sqrt 5 }}{4}\\
= \dfrac{4}{4} = 1\\
b.\dfrac{{3 + \sqrt 5 }}{{9 - 5}} - \dfrac{{\sqrt 5 + 1}}{{5 - 1}}\\
= \dfrac{{3 + \sqrt 5 - \sqrt 5 - 1}}{4} = \dfrac{1}{2}\\
d.\dfrac{{5 - 3\sqrt 2 + 5 + 3\sqrt 2 }}{{25 - 18}}\\
= \dfrac{{10}}{7}
\end{array}\)
