Question

rationalise the denominator of √6÷√3-√2​

Answer 1

Answer:

please refer to the attachment

I hope it would help you

thank you

Answer 2

$\dfrac{\sqrt{6}}{ \sqrt{3} \: - \: \sqrt{2} }$

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• We have to simply rationalize it.

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→ $\dfrac{\sqrt{6}}{ \sqrt{3} \: - \: \sqrt{2} }$

Rationalize it

→ $\dfrac{\sqrt{6}}{ \sqrt{3} \: - \: \sqrt{2} } \: \times \: \dfrac{ \sqrt{3} \: + \: \sqrt{2} }{ \sqrt{3} \: + \: \sqrt{2} }$

(a - b) (a + b) = a² - b²

→ $\dfrac{ \sqrt{6}(\sqrt{3} \: + \: \sqrt{2}) }{ (\sqrt{3})^{2} \: - \: (\sqrt{2})^{2} }$

→ $\dfrac{\sqrt{ 6}(\sqrt{3} \: + \: \sqrt{2}) }{ 3 \: - \: 2 }$

→ $\dfrac{\sqrt{18} \: + \: \sqrt{12}) }{ 1}$

→ $\sqrt{18}\:+\:\sqrt{12}$

→ $3\sqrt{2}\:+\:2\sqrt{3}$

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3√2 + 2√3

_____ [ ANSWER ]

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