Evaluate the following :
(32)25×(4)−12×(8)132−2÷(64)−13\dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{\dfrac{-1}{3}}}2−2÷(64)3−1(32)52×(4)−21×(8)31
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Given,
Simplifying the expression :
⇒[(2)5]25×[(2)2]−12×[(2)3]13(12)2÷(43)−13⇒(2)2×2−1×21(12)2÷4−1⇒(2)2×12×2(14)÷(14)⇒4×12×2(14)×4⇒4.\Rightarrow \dfrac{[(2)^5]^{\dfrac{2}{5}} \times [(2)^2]^{-\dfrac{1}{2}} \times {[(2)^3]^{\dfrac{1}{3}}}}{\Big(\dfrac{1}{2}\Big)^{2} ÷ (4^3)^{-\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{(2)^2 \times 2^{-1} \times 2^1}{\Big(\dfrac{1}{2}\Big)^{2} ÷ 4^{-1}} \\[1em] \Rightarrow \dfrac{(2)^2 \times \dfrac{1}{2} \times 2}{\Big(\dfrac{1}{4}\Big) ÷ \Big(\dfrac{1}{4}\Big)} \\[1em] \Rightarrow \dfrac{4 \times \dfrac{1}{2} \times 2}{\Big(\dfrac{1}{4}\Big) \times 4} \\[1em] \Rightarrow 4.⇒(21)2÷(43)−31[(2)5]52×[(2)2]−21×[(2)3]31⇒(21)2÷4−1(2)2×2−1×21⇒(41)÷(41)(2)2×21×2⇒(41)×44×21×2⇒4.
Hence, (32)25×(4)−12×(8)132−2÷(64)−13=4\dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{\dfrac{-1}{3}}} = 42−2÷(64)3−1(32)52×(4)−21×(8)31=4.
Answered By
(1681)−34×(499)32÷(343216)23\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}}(8116)−43×(949)23÷(216343)32
(64125)−23÷1(256625)14+(25643)0\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0(12564)−32÷(625256)411+(36425)0
(27)43+(32)0.8+(0.8)−1+(0.8)0(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0(27)34+(32)0.8+(0.8)−1+(0.8)0
[(27)−39−3]13\Big[\dfrac{(27)^{-3}}{9^{-3}}\Big]^{\dfrac{1}{3}}[9−3(27)−3]31
