Simplify the following:
(81)34−(132)−25+(8)13(12)−1(2)0(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0(81)43−(321)−52+(8)31(21)−1(2)0
23 Likes
Given,
⇒(81)34−(132)−25+(8)13(12)−1(2)0=(34)34−(32)25+(23)13(2)1.1=33−(25)25+2.2=27−22+4=27−4+4=27.\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0 = (3^4)^{\dfrac{3}{4}} - (32)^{\dfrac{2}{5}} + (2^3)^{\dfrac{1}{3}}(2)^1.1 \\[1em] = 3^3 - (2^5)^{\dfrac{2}{5}} + 2.2 \\[1em] = 27 - 2^2 + 4 \\[1em] = 27 - 4 + 4 = 27.⇒(81)43−(321)−52+(8)31(21)−1(2)0=(34)43−(32)52+(23)31(2)1.1=33−(25)52+2.2=27−22+4=27−4+4=27.
Hence, (81)34−(132)−25+(8)13(12)−1(2)0(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}}\Big(\dfrac{1}{2}\Big)^{-1}(2)^0(81)43−(321)−52+(8)31(21)−1(2)0 = 27.
Answered By
13 Likes
(9)52−3.(5)0−(181)−12(9)^{\dfrac{5}{2}} - 3.(5)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}(9)25−3.(5)0−(811)−21
1634+2(12)−1(3)016^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^01643+2(21)−1(3)0
(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
5n+3−6×5n+19×5n−22×5n\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 2^2 \times 5^n}9×5n−22×5n5n+3−6×5n+1
