Rationalize the denominator:
1(4+23)\dfrac{1}{(4 + 2\sqrt{3})}(4+23)1
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Rationalizing the denominator,
⇒1(4+23)×(4−23)(4−23)⇒(4−23)(4)2−(23)2⇒(4−23)16−12⇒2(2−3)4⇒(2−3)2\Rightarrow \dfrac{1}{(4 + 2\sqrt{3})} \times \dfrac{(4 - 2\sqrt{3})} {(4 - 2\sqrt{3})} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {(4)^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {16 - 12} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{3})} {4} \\[1em] \Rightarrow \dfrac{(2 - \sqrt{3})} {2} \\[1em]⇒(4+23)1×(4−23)(4−23)⇒(4)2−(23)2(4−23)⇒16−12(4−23)⇒42(2−3)⇒2(2−3)
Hence, on rationalizing 1(4+23)=(2−3)2\dfrac{1}{(4 + 2\sqrt{3})} = \dfrac{(2 - \sqrt{3})} {2}(4+23)1=2(2−3).
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3−13+1\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}3+13−1.
