Compute:
[23]−4×[278]−2\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2}[32]−4×[827]−2
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[23]−4×[278]−2=[32]4×[827]2=[3424]×[2333]2=[3424]×[2636]=[2636]×[2−43−4]=[26−436−4]=[2232]=[49]\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2}\\[1em] = \Big[\dfrac{3}{2}\Big]^4 \times \Big[\dfrac{8}{27}\Big]^2\\[1em] = \Big[\dfrac{3^4}{2^4}\Big] \times \Big[\dfrac{2^3}{3^3}\Big]^2\\[1em] = \Big[\dfrac{3^4}{2^4}\Big] \times \Big[\dfrac{2^6}{3^6}\Big]\\[1em] = \Big[\dfrac{2^6}{3^6}\Big] \times \Big[\dfrac{2^{-4}}{3^{-4}}\Big] \\[1em] = \Big[\dfrac{2^{6-4}}{3^{6-4}}\Big]\\[1em] = \Big[\dfrac{2^2}{3^2}\Big]\\[1em] = \Big[\dfrac{4}{9}\Big][32]−4×[827]−2=[23]4×[278]2=[2434]×[3323]2=[2434]×[3626]=[3626]×[3−42−4]=[36−426−4]=[3222]=[94]
Hence, [23]−4×[278]−2=[49]\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2} = \Big[\dfrac{4}{9}\Big][32]−4×[827]−2=[94]
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