Tính:
a) \(\sqrt {12} .\left( {\sqrt {12} + \sqrt 3 } \right);\)
b) \(\sqrt 8 .\left( {\sqrt {50} - \sqrt 2 } \right);\)
c) \({\left( {\sqrt 3 + \sqrt 2 } \right)^2} - 2\sqrt 6 .\)
Sử dụng kiến thức \(\sqrt A .\sqrt B = \sqrt {A.B} \)
a) \(\sqrt {12} .\left( {\sqrt {12} + \sqrt 3 } \right)\)
\(\begin{array}{l} = \sqrt {12} .\sqrt {12} + \sqrt {12} .\sqrt 3 \\ = \sqrt {{{12}^2}} .\sqrt {36} \\ = 12.6\\ = 72\end{array}\)
b) \(\sqrt 8 .\left( {\sqrt {50} - \sqrt 2 } \right)\)
\(\begin{array}{l} = \sqrt 8 .\sqrt {50} - \sqrt 8 .\sqrt 2 \\ = \sqrt {400} - \sqrt {16} \\ = 20 - 4\\ = 16\end{array}\)
c) \({\left( {\sqrt 3 + \sqrt 2 } \right)^2} - 2\sqrt 6 \)
\(\begin{array}{l} = {\sqrt 3 ^2} + 2.\sqrt 3 .\sqrt 2 + {\sqrt 2 ^2} - 2\sqrt 6 \\ = 3 + 2\sqrt 6 + 2 - 2\sqrt 6 \\ = 5\end{array}\)