Rationalise the denominator of 1/(2√5 + √3)​

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Question:

Rationalise the denominator of 1/(2√5 + √3)​

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Answer:[tex] \bf ➤ \underline{Solution-} \\ [/tex][tex] \sf \: \: \: \dfrac{1}{ 2 \sqrt{5} + \sqrt{3} } [/tex]On rationalising, [tex] \sf \implies\dfrac{1}{ 2 \sqrt{5} + \sqrt{3} } \times \dfrac{2 \sqrt{5} - \sqrt{3}}{2 \sqrt{5} - \sqrt{3}} [/tex]Combine the fractions, [tex] \sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} + \sqrt{3})(2 \sqrt{5} - \sqrt{3})} [/tex]We know that, [tex] \sf \implies (a + b)(a - b) = (a)^{2} - (b)^{2} [/tex]So, [tex] \sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{(2 \sqrt{5} )^{2} - (\sqrt{3})^{2} } [/tex][tex] \sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{20 - 3 } [/tex][tex] \sf \implies \dfrac{1(2 \sqrt{5} - \sqrt{3})}{17 } [/tex][tex] \sf \implies \dfrac{2 \sqrt{5} - \sqrt{3}}{17 } [/tex]Hence, On rationalising we got, [tex] \bf \implies\dfrac{2 \sqrt{5} - \sqrt{3}}{17 } [/tex]

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