Evaluate:
(3−1×9−1)÷3−2(3^{-1} \times 9^{-1}) ÷ 3^{-2}(3−1×9−1)÷3−2
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As we know, for any non-zero rational number a
a−n=1ana^{-n} = \dfrac{1}{a^n}a−n=an1 and an=1a−na^{n} = \dfrac{1}{a^{-n}}an=a−n1.
Hence,
(3−1×9−1)÷3−2=(131×191)÷(13)2=(13×19)÷(1×13×3)=(1×13×9)÷(19)=(127)÷(19)=(127)×(91)=1×927×1=927=13(3^{-1} \times 9^{-1}) ÷ 3^{-2}\\[1em] = \Big(\dfrac{1}{3}^1 \times \dfrac{1}{9}^1\Big) ÷ \Big(\dfrac{1}{3}\Big)^2\\[1em] = \Big(\dfrac{1}{3} \times \dfrac{1}{9}\Big) ÷ \Big(\dfrac{1 \times 1}{3 \times 3}\Big)\\[1em] = \Big(\dfrac{1 \times 1}{3 \times 9}\Big) ÷ \Big(\dfrac{1}{9}\Big)\\[1em] = \Big(\dfrac{1}{27}\Big) ÷ \Big(\dfrac{1}{9}\Big)\\[1em] = \Big(\dfrac{1}{27}\Big) \times \Big(\dfrac{9}{1}\Big)\\[1em] = \dfrac{1 \times 9}{27 \times 1}\\[1em] = \dfrac{9}{27}\\[1em] = \dfrac{1}{3}(3−1×9−1)÷3−2=(311×911)÷(31)2=(31×91)÷(3×31×1)=(3×91×1)÷(91)=(271)÷(91)=(271)×(19)=27×11×9=279=31
Hence, (3−1×9−1)÷3−2=13(3^{-1} \times 9^{-1}) ÷ 3^{-2} = \dfrac{1}{3}(3−1×9−1)÷3−2=31
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