$\dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{6}}$ equals

$\dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{6}}$ equals
  1. $\dfrac{\sqrt{6}}{3}$
  2. $\sqrt{6}$
  3. $\dfrac{2}{6}$
  4. $\dfrac{2}{\sqrt{6}}$

Solution

$\dfrac{2}{\sqrt{6}} \cdot \dfrac{\sqrt{6}}{\sqrt{6}} = \dfrac{2\sqrt{6}}{6} = \dfrac{\sqrt{6}}{3}$.

Asked in: MH-SSC-9

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