Simplify the following:
(278)23−(14)−2+50\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0(827)32−(41)−2+50
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Given,
⇒(278)23−(14)−2+50=(3323)23−42+1=33×2323×23−16+1=3222−15=94−15=9−604=−514=−1234.\Rightarrow \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = \Big(\dfrac{3^3}{2^3}\Big)^{\dfrac{2}{3}} - 4^2 + 1 \\[1em] = \dfrac{3^{3 \times \dfrac{2}{3}}}{2^{3 \times \dfrac{2}{3}}} - 16 + 1 \\[1em] = \dfrac{3^2}{2^2} - 15 \\[1em] = \dfrac{9}{4} - 15 \\[1em] = \dfrac{9 - 60}{4} \\[1em] = -\dfrac{51}{4} = -12\dfrac{3}{4}.⇒(827)32−(41)−2+50=(2333)32−42+1=23×3233×32−16+1=2232−15=49−15=49−60=−451=−1243.
Hence, (278)23−(14)−2+50=−1234.\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = -12\dfrac{3}{4}.(827)32−(41)−2+50=−1243.
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