Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {\sqrt 6 + \sqrt 2 } \right).\left( {\sqrt 3 - 2} \right).\sqrt {2 + \sqrt 3 } \\
= \sqrt 2 .\left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 2} \right).\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 2} \right).\sqrt {4 + 2\sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 2} \right).\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 2} \right).\left( {\sqrt 3 + 1} \right)\\
= {\left( {\sqrt 3 + 1} \right)^2}.\left( {\sqrt 3 - 2} \right)\\
= \left( {4 + 2\sqrt 3 } \right)\left( {\sqrt 3 - 2} \right)\\
= 4\sqrt 3 - 8 + 2.3 - 4\sqrt 3 \\
= - 2\\
B = \left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} - \sqrt 6 } \right).\sqrt {4 - \sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right).\sqrt 2 .\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {4 - \sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {8 - 2\sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\\
= \left( {4 + \sqrt {15} } \right).{\left( {\sqrt 5 - \sqrt 3 } \right)^2}\\
= \left( {4 + \sqrt {15} } \right).\left( {8 - 2\sqrt {15} } \right)\\
= \left( {4 + \sqrt {15} } \right).2.\left( {4 - \sqrt {15} } \right)\\
= 2.\left( {{4^2} - {{\sqrt {15} }^2}} \right)\\
= 2
\end{array}\)
