Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{1}{3}\sqrt {27} - \sqrt {48} + 9\sqrt {\dfrac{1}{3}} + \sqrt {12} \\
= \dfrac{1}{3}.\sqrt {{3^2}.3} - \sqrt {{4^2}.3} + 3.\dfrac{3}{{\sqrt 3 }} + \sqrt {{2^2}.3} \\
= \dfrac{1}{3}.3.\sqrt 3 - 4.\sqrt 3 + 3.\sqrt 3 + 2\sqrt 3 \\
= \sqrt 3 - 4\sqrt 3 + 3\sqrt 3 + 2\sqrt 3 \\
= 2\sqrt 3 \\
B = \left( {3 + \sqrt 5 } \right).\sqrt {14 - 6\sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\sqrt {9 - 2.3.\sqrt 5 + 5} \\
= \left( {3 + \sqrt 5 } \right).\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} \\
= \left( {3 + \sqrt 5 } \right).\left( {3 - \sqrt 5 } \right)\\
= {3^2} - {\sqrt 5 ^2}\\
= 9 - 5\\ = 4\\
C = \dfrac{{\sqrt {10} + 5\sqrt 2 }}{{\sqrt 5 + 1}} - 6\sqrt {\dfrac{5}{2}} + \dfrac{{12}}{{4 - \sqrt {10} }}\\
= \dfrac{{\sqrt 5 .\sqrt 2 + {{\sqrt 5 }^2}.\sqrt 2 }}{{\sqrt 5 + 1}} - 3.\dfrac{2}{{\sqrt 2 }}.\sqrt 5 + \dfrac{{12\left( {4 + \sqrt {10} } \right)}}{{\left( {4 - \sqrt {10} } \right)\left( {4 + \sqrt {10} } \right)}}\\
= \dfrac{{\sqrt 5 .\sqrt 2 .\left( {1 + \sqrt 5 } \right)}}{{\sqrt 5 + 1}} - 3.\sqrt 2 .\sqrt 5 + \dfrac{{12.\left( {4 + \sqrt {10} } \right)}}{{{4^2} - 10}}\\
= \sqrt {10} - 3.\sqrt {10} + 2.\left( {4 + \sqrt {10} } \right)\\
= \sqrt {10} - 3\sqrt {10} + 8 + 2\sqrt {10} \\
= 8
\end{array}\)
