Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
\left( {4\sqrt {27} - 2\sqrt {48} - 5\sqrt {75} } \right):2\sqrt 3 \\
= \left( {4.\sqrt {{3^2}.3} - 2.\sqrt {{4^2}.3} - 5.\sqrt {{5^2}.3} } \right):2\sqrt 3 \\
= \left( {4.3\sqrt 3 - 2.4\sqrt 3 - 5.5\sqrt 3 } \right):2\sqrt 3 \\
= \left( {12\sqrt 3 - 8\sqrt 3 - 25\sqrt 3 } \right):2\sqrt 3 \\
= \left( { - 21\sqrt 3 } \right):2\sqrt 3 \\
= - \dfrac{{21}}{2}\\
c,\\
\dfrac{{\sqrt 3 }}{{1 - \sqrt {\sqrt 3 + 1} }} + \dfrac{{\sqrt 3 }}{{1 + \sqrt {\sqrt 3 + 1} }}\\
= \dfrac{{\sqrt 3 .\left( {1 + \sqrt {\sqrt 3 + 1} } \right) + \sqrt 3 .\left( {1 - \sqrt {\sqrt 3 + 1} } \right)}}{{\left( {1 - \sqrt {\sqrt 3 + 1} } \right).\left( {1 + \sqrt {\sqrt 3 + 1} } \right)}}\\
= \dfrac{{\sqrt 3 + \sqrt 3 .\sqrt {\sqrt 3 + 1} + \sqrt 3 - \sqrt 3 .\sqrt {\sqrt 3 + 1} }}{{{1^2} - {{\sqrt {\sqrt 3 + 1} }^2}}}\\
= \dfrac{{2\sqrt 3 }}{{1 - \left( {\sqrt 3 + 1} \right)}}\\
= \dfrac{{2\sqrt 3 }}{{ - \sqrt 3 }}\\
= - 2\\
d,\\
\dfrac{3}{{2 + \sqrt 3 }} + \dfrac{{13}}{{4 - \sqrt 3 }} + \dfrac{6}{{\sqrt 3 }}\\
= \dfrac{{3.\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} + \dfrac{{13\left( {4 + \sqrt 3 } \right)}}{{\left( {4 - \sqrt 3 } \right)\left( {4 + \sqrt 3 } \right)}} + \dfrac{{2.3}}{{\sqrt 3 }}\\
= \dfrac{{3.\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}} + \dfrac{{13\left( {4 + \sqrt 3 } \right)}}{{{4^2} - 3}} + \dfrac{{2.{{\sqrt 3 }^2}}}{{\sqrt 3 }}\\
= \dfrac{{3\left( {2 - \sqrt 3 } \right)}}{1} + \dfrac{{13.\left( {4 + \sqrt 3 } \right)}}{{13}} + 2\sqrt 3 \\
= 6 - 3\sqrt 3 + 4 + \sqrt 3 + 2\sqrt 3 \\
= 10
\end{array}\)
Em xem lại đề câu a nhé!
