Rationalise the denominator and simplify :
125−3\dfrac{1}{2\sqrt5 - \sqrt3}25−31
Since, the denominator = 25−32\sqrt5 - \sqrt325−3, its rationalizing factor = 25+32\sqrt5 + \sqrt325+3.
125−3=125−3×25+325+3=1(25+3)(25)2−(3)2=25+320−3=25+317\dfrac{1}{2\sqrt5 - \sqrt3} = \dfrac{1}{2\sqrt5 - \sqrt3} \times \dfrac{2\sqrt5 + \sqrt3}{2\sqrt5 + \sqrt3}\\[1em] = \dfrac{1(2\sqrt5 + \sqrt3)}{(2\sqrt5)^2 - (\sqrt3)^2} \\[1em] = \dfrac{2\sqrt5 + \sqrt3}{20 - 3} \\[1em] = \dfrac{2\sqrt5 + \sqrt3}{17}25−31=25−31×25+325+3=(25)2−(3)21(25+3)=20−325+3=1725+3
Hence, 125−3=25+317\dfrac{1}{2\sqrt5 - \sqrt3} = \dfrac{2\sqrt5 + \sqrt3}{17}25−31=1725+3.
Answered By
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5+35−3\dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3}5−35+3
Simplify :
4+54−5+4−54+5\dfrac{4 + \sqrt5}{4 - \sqrt5} + \dfrac{4 - \sqrt5}{4 + \sqrt5}4−54+5+4+54−5