Evaluate:
[{(−13)−2}2]−1\Big[\Big{\Big(-\dfrac{1}{3}\Big)^{-2}\Big}^2\Big]^{-1}[{(−31)−2}2]−1
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[{(−13)−2}2]−1=[{(−3)2}2]−1=[{9}2]−1=[81]−1=181\Big[\Big{\Big(-\dfrac{1}{3}\Big)^{-2}\Big}^2\Big]^{-1}\\[1em] = [{(-3)^2}^2]^{-1}\\[1em] = [{9}^2]^{-1}\\[1em] = [81]^{-1}\\[1em] = \dfrac{1}{81}[{(−31)−2}2]−1=[{(−3)2}2]−1=[{9}2]−1=[81]−1=811
[{(−13)−2}2]−1=181\Big[\Big{\Big(-\dfrac{1}{3}\Big)^{-2}\Big}^2\Big]^{-1} = \dfrac{1}{81}[{(−31)−2}2]−1=811
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(80+2−1)×32(8^0 + 2^{-1})\times 3^2(80+2−1)×32
[(16)−1−(15)−1]−2\Big[\Big(\dfrac{1}{6}\Big)^{-1} - \Big(\dfrac{1}{5}\Big)^{-1}\Big]^{-2}[(61)−1−(51)−1]−2
5n+2−5n+15n+1\dfrac{5^{n+2}-5^{n+1}}{5^{n+1}}5n+15n+2−5n+1
Find the value of x; if:
1(125)x−7=52x−1\dfrac{1}{(125)^{x-7}}=5^{2x-1}(125)x−71=52x−1
