Simplify the following:
[(64)23.2−2÷80]−12\Big[(64)^{\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}}[(64)32.2−2÷80]−21
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Given,
⇒[(64)23.2−2÷80]−12=[[(4)3]23×(12)2÷1]−12=[(42)×(14)]−12=[16×14]−12=[4]−12=(14)12=12=12.\Rightarrow \Big[(64)^{\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}} = \Big[[(4)^3]^{\dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^2 ÷ 1 \Big]^{-\dfrac{1}{2}} \\[1em] = \Big[(4^2) \times \Big(\dfrac{1}{4}\Big)\Big]^{-\dfrac{1}{2}} \\[1em] = \Big[16 \times \dfrac{1}{4}\Big]^{-\dfrac{1}{2}} \\[1em] = [4]^{-\dfrac{1}{2}} \\[1em] = \Big(\dfrac{1}{4}\Big)^{\dfrac{1}{2}} = \dfrac{1}{2} \\[1em] = \dfrac{1}{2}.⇒[(64)32.2−2÷80]−21=[[(4)3]32×(21)2÷1]−21=[(42)×(41)]−21=[16×41]−21=[4]−21=(41)21=21=21.
Hence, [(64)−23.2−2÷80]−12=12\Big[(64)^{-\dfrac{2}{3}}.2^{-2} ÷ 8^0\Big]^{-\dfrac{1}{2}} = \dfrac{1}{2}[(64)−32.2−2÷80]−21=21.
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