Question

rationalise the denominator of √6/√2+ √3

Answer 1

$\frac{\sqrt{6}}{(\sqrt{2} + \sqrt{3})}$

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

$= \frac{\sqrt{6\times3} - \sqrt{6\times 2}}{(\sqrt{3})^{2} - (\sqrt{2})^{2}}$

$= \frac{\sqrt{2\times 2 \times 3 } - \sqrt{2 \times 3\times 3}}{3-2}$

$= 2\sqrt{3} - 3\sqrt{2}$

Therefore.,

$\red {\frac{\sqrt{6}}{(\sqrt{2} + \sqrt{3})}}\green {= 2\sqrt{3} - 3\sqrt{2}}$

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