Evaluate the following :
[(64)23×2−2÷70]−12\Big[(64)^{\dfrac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\dfrac{1}{2}}[(64)32×2−2÷70]−21
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Given,
Simplifying the expression :
⇒[[(4)3]23×(12)2÷1]−12⇒[(4)3×23×(12)2÷1]−12⇒[(4)2×(14)÷1]−12⇒[16×(14)]−12⇒(4)−12⇒(14)12⇒[(12)2]12⇒12.\Rightarrow \Big[[(4)^3]^{\dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^{2} ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[(4)^{3 \times \dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^{2} ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[(4)^2 \times \Big(\dfrac{1}{4}\Big) ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[16 \times \Big(\dfrac{1}{4}\Big)\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow (4)^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big)^{\dfrac{1}{2}} \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{2}\Big)^2\Big]^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{1}{2}.⇒[[(4)3]32×(21)2÷1]−21⇒[(4)3×32×(21)2÷1]−21⇒[(4)2×(41)÷1]−21⇒[16×(41)]−21⇒(4)−21⇒(41)21⇒[(21)2]21⇒21.
Hence, [(64)23×2−2÷70]−12=12\Big[(64)^{\dfrac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\dfrac{1}{2}} = \dfrac{1}{2}[(64)32×2−2÷70]−21=21.
Answered By
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