Question

Rationalise the denominator of 1/√3-√2​

Answer 1

Step-by-step explanation:

$\frac{1}{\sqrt{3}-\sqrt{2} } =\frac{1}{\sqrt{3}-\sqrt{2} }*\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\=\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2} )^{2} } = \frac{\sqrt{3}+\sqrt{2}}{3-2} =\frac{\sqrt{3}+\sqrt{2}}{1} =\sqrt{3}+\sqrt{2}$

Answer = √3+√2

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Answer 2

S O L U T I O N :

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$\sf : \implies \: \dfrac{1}{ (\sqrt{3} - \sqrt{2} ) }$

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$\sf \underline{Multiply \: numerator \: and \: denominator \: with \: \bold{( \sqrt{3} + \sqrt{2} )}.}$

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$\sf : \implies \: \dfrac{1}{ (\sqrt{3} - \sqrt{2}) } \times \dfrac{ (\sqrt{3} + \sqrt{2}) }{ (\sqrt{3} + \sqrt{2} ) }$

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$\sf {\underline{As \: we \: know \: that \: \bold{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}$

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

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

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

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$\sf : \implies \: \bold{( \sqrt{3} + \sqrt{2} )}$

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$\sf\therefore{\underline{Hence, \: the \: answer \: is \: \bold{( \sqrt{3} + \sqrt{2}) }.}}$