Compute:
(27)23÷(8116)−14(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}}(27)32÷(1681)−41
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(27)23÷(8116)−14=(33)23÷(3424)−14=(3)3×23÷(32)−4×14=(3)2÷(32)−1=(3)2÷(23)=(3)2×(32)=9×(32)=9×32=272=1312(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}}\\[1em] = (3^3)^{\dfrac{2}{3}} ÷ \Big(\dfrac{3^4}{2^4}\Big)^{-\dfrac{1}{4}}\\[1em] = (3)^{3\times \dfrac{2}{3}} ÷ \Big(\dfrac{3}{2}\Big)^{-4\times\dfrac{1}{4}}\\[1em] = (3)^2 ÷ \Big(\dfrac{3}{2}\Big)^{-1}\\[1em] = (3)^2 ÷ \Big(\dfrac{2}{3}\Big)\\[1em] = (3)^2 \times \Big(\dfrac{3}{2}\Big)\\[1em] = 9 \times \Big(\dfrac{3}{2}\Big)\\[1em] = \dfrac{9 \times 3}{2}\\[1em] = \dfrac{27}{2}\\[1em] = 13\dfrac{1}{2}(27)32÷(1681)−41=(33)32÷(2434)−41=(3)3×32÷(23)−4×41=(3)2÷(23)−1=(3)2÷(32)=(3)2×(23)=9×(23)=29×3=227=1321
(27)23÷(8116)−14=1312(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}} = 13\dfrac{1}{2}(27)32÷(1681)−41=1321
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(243)25÷(32)−25(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}(243)52÷(32)−52
(−3)4−(34)0×(−2)5÷(64)23(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}}(−3)4−(43)0×(−2)5÷(64)32
Simplify:
843+2532−(127)−238^{\dfrac{4}{3}} + 25^{\dfrac{3}{2}} - \Big(\dfrac{1}{27}\Big)^{-\dfrac{2}{3}}834+2523−(271)−32
[(64)−2]−3÷[{(−8)2}3]2[(64)^{-2}]^{-3} ÷ [{(-8)^2}^3]^2[(64)−2]−3÷[{(−8)2}3]2
