Compute:
(243)25÷(32)−25(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}(243)52÷(32)−52
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(243)25÷(32)−25=(35)25÷(25)−25=(3)5×25÷(2)−5×25=(3)2÷(2)−2=(3)2×(12)−2=(3)2×212=(9)×(4)=36(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}\\[1em] = (3^5)^{\dfrac{2}{5}} ÷ (2^5)^{-\dfrac{2}{5}}\\[1em] = (3)^{5\times\dfrac{2}{5}} ÷ (2)^{-5\times\dfrac{2}{5}}\\[1em] = (3)^{2} ÷ (2)^{-2}\\[1em] = (3)^{2} \times \Big(\dfrac{1}{2}\Big)^{-2}\\[1em] = (3)^{2} \times \dfrac{2}{1}^{2}\\[1em] = (9) \times (4)\\[1em] = 36(243)52÷(32)−52=(35)52÷(25)−52=(3)5×52÷(2)−5×52=(3)2÷(2)−2=(3)2×(21)−2=(3)2×122=(9)×(4)=36
(243)25÷(32)−25=36(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}} = 36(243)52÷(32)−52=36
Answered By
[132]−25\Big[\dfrac{1}{32}\Big]^{-\dfrac{2}{5}}[321]−52
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