Rationalise the denominator and simplify :
25+3\dfrac{2}{\sqrt5 + \sqrt3}5+32
Since, the denominator = 5+3\sqrt5 + \sqrt35+3, its rationalizing factor = 5−3\sqrt5 - \sqrt35−3.
25+3=25+3×5−35−3=2(5−3)(5)2−(3)2=2(5−3)5−3=2(5−3)2=(5−3)\dfrac{2}{\sqrt5 + \sqrt3} = \dfrac{2}{\sqrt5 + \sqrt3} \times \dfrac{\sqrt5 - \sqrt3}{\sqrt5 - \sqrt3}\\[1em] = \dfrac{2(\sqrt5 - \sqrt3)}{(\sqrt5)^2 - (\sqrt3)^2} \\[1em] = \dfrac{2(\sqrt5 - \sqrt3)}{5 - 3} \\[1em] = \dfrac{2(\sqrt5 - \sqrt3)}{2}\\[1em] = (\sqrt5 - \sqrt3)5+32=5+32×5−35−3=(5)2−(3)22(5−3)=5−32(5−3)=22(5−3)=(5−3)
Hence, 25+3=(5−3)\dfrac{2}{\sqrt5 + \sqrt3} = (\sqrt5 - \sqrt3)5+32=(5−3).
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