Question

Solve questions no. 6​

Answer 1

To solve :-

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

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Solution :-

We have to Rationalise the given fraction by multiplying it by the conjugate of its denominator. Conjugate of 3 - √6 is 3 + √6.

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

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

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Solving numerator

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${ \implies \dfrac{6 \sqrt{3} + 2\sqrt{18} }{(3 - \sqrt{6}) (3 + \sqrt{6}) }}$

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${ \implies \dfrac{6 \sqrt{3} + 2\sqrt{3 \times 3 \times 2} }{(3 - \sqrt{6}) (3 + \sqrt{6}) }}$

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

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Apply identity in numerator (A - B) (A + B) = A² - B²

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

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${ \implies \dfrac{6 \sqrt{3} + 6\sqrt{2} }{9 - 6 }}$

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${ \implies \dfrac{6 \sqrt{3} + 6\sqrt{2} }{3 }}$

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Taking 6 common in numerator

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${ \implies \dfrac{6( \sqrt{3} + \sqrt{2} )}{3 }}$

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${ \implies \dfrac{ \not6( \sqrt{3} + \sqrt{2} )}{ \not3 }}$

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${ \implies \dfrac{2( \sqrt{3} + \sqrt{2} )}{1 }}$

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$\implies 2( \sqrt{3} + \sqrt{2})$

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This is the final answer.