Question

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Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)

\(=\dfrac{3}{2}\sqrt{6}+2\cdot\sqrt{\dfrac{6}{9}}-4\cdot\sqrt{\dfrac{6}{4}}\)

\(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)

\(=\dfrac{1}{6}\sqrt{6}\)

2: \(2\sqrt{48}+6\sqrt{\dfrac{1}{3}}-4\sqrt{12}\)

\(=2\cdot4\sqrt{3}+\dfrac{6}{\sqrt{3}}-4\cdot2\sqrt{3}\)

\(=8\sqrt{3}-8\sqrt{3}+2\sqrt{3}=2\sqrt{3}\)

8: \(\sqrt{4-\sqrt{7}}\cdot\sqrt{4+\sqrt{7}}\)

\(=\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

\(=\sqrt{16-7}=\sqrt{9}=3\)

9: \(\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2-\sqrt{2}}\)

\(=\sqrt{2}\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)

\(=\sqrt{2}\cdot\sqrt{4-2}\)

\(=\sqrt{2\cdot2}=2\)