Question

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

8) ( 1+ \(\sqrt{2}\)  + \(\sqrt{3}\) ) ( 1 + \(\sqrt{2}\) - \(\sqrt{3}\) )

Answer 1

7: Ta có: \(\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right):\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{\sqrt{6}}{3}\right)\cdot\sqrt{6}\)

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

8: ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)

\(=\left(\sqrt{2}+1\right)^2-3\)

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

\(=2\sqrt{2}\)

Answer 2

7)\(\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right):\dfrac{1}{\sqrt{6}}=\dfrac{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}{\sqrt{6}}.\sqrt{6}=3-2=1\)

8) \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)=\left(1+\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2=1+2+2\sqrt{2}-3=2\sqrt{2}\)