Rationalise the denominator and simplify :
5+35−3\dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3}5−35+3
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Since, the denominator = 5−3\sqrt5 - \sqrt35−3, its rationalizing factor = 5+3\sqrt5 + \sqrt35+3.
5+35−3=5+35−3×5+35+3=(5+3)2(5)2−(3)2=5+3+2155−3=8+2152=4+15\dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} = \dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} \times \dfrac{\sqrt5 + \sqrt3}{\sqrt5 + \sqrt3}\\[1em] = \dfrac{(\sqrt5 + \sqrt3)^2}{(\sqrt5)^2 - (\sqrt3)^2} \\[1em] = \dfrac{5 + 3 + 2\sqrt{15}}{5 - 3} \\[1em] = \dfrac{8 + 2\sqrt{15}}{2}\\[1em] = 4 + \sqrt{15}5−35+3=5−35+3×5+35+3=(5)2−(3)2(5+3)2=5−35+3+215=28+215=4+15
Hence, 5+35−3=4+15\dfrac{\sqrt5 + \sqrt3}{\sqrt5 - \sqrt3} = 4 + \sqrt{15}5−35+3=4+15.
Answered By
1223+6\dfrac{12\sqrt2}{\sqrt3 + \sqrt6}3+6122
125−3\dfrac{1}{2\sqrt5 - \sqrt3}25−31
Simplify :
4+54−5+4−54+5\dfrac{4 + \sqrt5}{4 - \sqrt5} + \dfrac{4 - \sqrt5}{4 + \sqrt5}4−54+5+4+54−5
35−3+25+3\dfrac{3}{5 - \sqrt3} + \dfrac{2}{5 + \sqrt3}5−33+5+32
