Compute:
(125)−23÷(8)23(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}}(125)−32÷(8)32
1 Like
(125)−23÷(8)23=(53)−23÷(23)23=(5)−3×23÷(2)3×23=(5)−2÷(2)2=(5)−2×(12)2=(5)−2×14=(15)2×14=125×14=1100(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}}\\[1em] = (5^3)^{-\dfrac{2}{3}} ÷ (2^3)^{\dfrac{2}{3}}\\[1em] = (5)^{-3\times\dfrac{2}{3}} ÷ (2)^{3\times\dfrac{2}{3}}\\[1em] = (5)^{-2} ÷ (2)^{2}\\[1em] = (5)^{-2} \times \Big(\dfrac{1}{2}\Big)^{2}\\[1em] = (5)^{-2} \times \dfrac{1}{4}\\[1em] = \Big(\dfrac{1}{5}\Big)^2 \times \dfrac{1}{4}\\[1em] = \dfrac{1}{25} \times \dfrac{1}{4}\\[1em] = \dfrac{1}{100}(125)−32÷(8)32=(53)−32÷(23)32=(5)−3×32÷(2)3×32=(5)−2÷(2)2=(5)−2×(21)2=(5)−2×41=(51)2×41=251×41=1001
(125)−23÷(8)23=1100(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}} = \dfrac{1}{100}(125)−32÷(8)32=1001
Answered By
[2764]−23\Big[\dfrac{27}{64}\Big]^{-\dfrac{2}{3}}[6427]−32
[132]−25\Big[\dfrac{1}{32}\Big]^{-\dfrac{2}{5}}[321]−52
(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