Question

Tính:

a) \(\sqrt{12}.\left(\sqrt{12}+\sqrt{3}\right)\);                b) \(\sqrt{8}.\left(\sqrt{50}-\sqrt{2}\right)\);                    c) \(\left(\sqrt{3}+\sqrt{2}\right)^2-2\sqrt{6}\).

Answer 1

a) \(\sqrt {12} .\left( {\sqrt {12}  + \sqrt 3 } \right)\)

\(\begin{array}{l} = \sqrt {12} .\sqrt {12}  + \sqrt {12} .\sqrt 3 \\ = \sqrt {{{12}^2}} +\sqrt {36} \\ = 12+6\\ = 18\end{array}\)

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) \({\left( {\sqrt 3  + \sqrt 2 } \right)^2} - 2\sqrt 6 \)

\(\begin{array}{l} = {\sqrt 3 ^2} + 2.\sqrt 3 .\sqrt 2  + {\sqrt 2 ^2} - 2\sqrt 6 \\ = 3 + 2\sqrt 6  + 2 - 2\sqrt 6 \\ = 5\end{array}\)