Rationalise the denominator and simplify :
1223+6\dfrac{12\sqrt2}{\sqrt3 + \sqrt6}3+6122
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Since, the denominator = 3+6\sqrt3 + \sqrt63+6, its rationalizing factor = 3−6\sqrt3 - \sqrt63−6.
1223+6=1223+6×3−63−6=12(6−23)(3)2−(6)2=12(6−23)3−6=12(6−23)−3=−4(6−23)=83−46\dfrac{12\sqrt2}{\sqrt3 + \sqrt6} = \dfrac{12\sqrt2}{\sqrt3 + \sqrt6} \times \dfrac{\sqrt3 - \sqrt6}{\sqrt3 - \sqrt6}\\[1em] = \dfrac{12(\sqrt6 - 2\sqrt3)}{(\sqrt3)^2 - (\sqrt6)^2} \\[1em] = \dfrac{12(\sqrt6 - 2\sqrt3)}{3 - 6} \\[1em] = \dfrac{12(\sqrt6 - 2\sqrt3)}{-3}\\[1em] = -4(\sqrt6 - 2\sqrt3)\\[1em] = 8\sqrt3 - 4\sqrt63+6122=3+6122×3−63−6=(3)2−(6)212(6−23)=3−612(6−23)=−312(6−23)=−4(6−23)=83−46
Hence, 1223+6=83−46\dfrac{12\sqrt2}{\sqrt3 + \sqrt6} = 8\sqrt3 - 4\sqrt63+6122=83−46.
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