Math, asked by danieleldo077, 2 days ago

rationalise 4 root 3 /root 6 + root 2​

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Answered by ramaraogudibandla
0

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Answered by KingdomOfNepal
2

Given :

\dfrac{4\sqrt{3}}{\sqrt{6}\: +\: \sqrt{2}}

Rationalizing the denominator :

\dfrac{4\sqrt{3}}{\sqrt{6}\: +\: \sqrt{2}}\: \times\: \dfrac{\sqrt{6}\: -\: \sqrt{2}}{\sqrt{6}\: -\: \sqrt{2}}

=\: \dfrac{4\sqrt{3}(\sqrt{6}\: -\: \sqrt{2})}{(\sqrt{6})^2\: -\: (\sqrt{2})^2}

(since (a+b)(a-b) = a² - b²)

=\: \dfrac{4\sqrt{18}\: - 4\sqrt{6}}{6\: -\: 2}

=\: \dfrac{\cancel4\:(\sqrt{18}\: -\: \sqrt{6})}{\cancel4}

= √18 - √6

= 3√2 - √6

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