Evaluate:
14+(0.01)−12×(5)−(27)23\sqrt\dfrac{1}{4} + (0.01)^{-\dfrac{1}{2}} \times (5) - (27)^\dfrac{2}{3}41+(0.01)−21×(5)−(27)32
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14+(0.01)−12×5−(27)23=12+(10.01)12×5−(33)23=12+(1001)12×5−(3)2×33=12+100×5−(3)2=12+10×5−9=12+50−9=12+41=1+822=832=4112\sqrt\dfrac{1}{4} + (0.01)^{-\dfrac{1}{2}} \times 5 - (27)^\dfrac{2}{3}\\[1em] = \dfrac{1}{2} + \Big(\dfrac{1}{0.01}\Big)^{\dfrac{1}{2}} \times 5 - (3^3)^\dfrac{2}{3}\\[1em] = \dfrac{1}{2} + \Big(\dfrac{100}{1}\Big)^{\dfrac{1}{2}} \times 5 - (3)^\dfrac{2 \times 3}{3}\\[1em] = \dfrac{1}{2} + \sqrt{100} \times 5 - (3)^2\\[1em] = \dfrac{1}{2} + 10 \times 5 - 9\\[1em] = \dfrac{1}{2} + 50 - 9\\[1em] = \dfrac{1}{2} + 41\\[1em] = \dfrac{1 + 82}{2}\\[1em] = \dfrac{83}{2}\\[1em] = 41\dfrac{1}{2}41+(0.01)−21×5−(27)32=21+(0.011)21×5−(33)32=21+(1100)21×5−(3)32×3=21+100×5−(3)2=21+10×5−9=21+50−9=21+41=21+82=283=4121
Hence, 14+(0.01)−12×(5)−(27)23=4112\sqrt\dfrac{1}{4} + (0.01)^{-\dfrac{1}{2}} \times (5) - (27)^\dfrac{2}{3} = 41\dfrac{1}{2}41+(0.01)−21×(5)−(27)32=4121.
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