Rationalize the denominator:
1(25−3)\dfrac{1}{(2\sqrt{5} - \sqrt{3})}(25−3)1.
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Rationalizing the denominator,
⇒1(25−3)×(25+3)(25+3)⇒(25+3)(25)2−(3)2⇒(25+3)20−3⇒(25+3)17\Rightarrow \dfrac{1}{(2\sqrt{5} - \sqrt{3})} \times \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {20 - 3} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {17} \\[1em]⇒(25−3)1×(25+3)(25+3)⇒(25)2−(3)2(25+3)⇒20−3(25+3)⇒17(25+3)
Hence, on rationalizing 1(25−3)=(25+3)17\dfrac{1}{(2\sqrt{5} - \sqrt{3})} = \dfrac{(2\sqrt{5} + \sqrt{3})}{17}(25−3)1=17(25+3).
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3−13+1\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}3+13−1.
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1(1+5+3)\dfrac{1}{(1 + \sqrt{5} + \sqrt{3})}(1+5+3)1.
2(2+3−5)\dfrac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})}(2+3−5)2.