Evaluate:
(3−1×4−1)÷6−1(3^{-1} \times 4^{-1}) ÷ 6^{-1}(3−1×4−1)÷6−1
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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,
(3−1×4−1)÷6−1=(131×141)÷(16)1=(1×13×4)÷(16)=(112)÷(16)=(112)×(61)=(1×612×1)=612=12(3^{-1} \times 4^{-1}) ÷ 6^{-1}\\[1em] = \Big(\dfrac{1}{3}^1 \times \dfrac{1}{4}^1\Big) ÷ \Big(\dfrac{1}{6}\Big)^1\\[1em] = \Big(\dfrac{1 \times 1}{3 \times 4}\Big) ÷ \Big(\dfrac{1}{6}\Big)\\[1em] = \Big(\dfrac{1}{12}\Big) ÷ \Big(\dfrac{1}{6}\Big)\\[1em] = \Big(\dfrac{1}{12}\Big) \times \Big(\dfrac{6}{1}\Big)\\[1em] = \Big(\dfrac{1 \times 6}{12 \times 1}\Big)\\[1em] = \dfrac{6}{12}\\[1em] = \dfrac{1}{2}(3−1×4−1)÷6−1=(311×411)÷(61)1=(3×41×1)÷(61)=(121)÷(61)=(121)×(16)=(12×11×6)=126=21
Hence, (3−1×4−1)÷6−1=12(3^{-1} \times 4^{-1}) ÷ 6^{-1} = \dfrac{1}{2}(3−1×4−1)÷6−1=21
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(−5)5×(−5)−3(-5)^5 \times (-5)^{-3}(−5)5×(−5)−3 is equal to:
15\dfrac{1}{5}51
5
-25
25
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(3−1÷4−1)2(3^{-1} ÷ 4^{-1})^2(3−1÷4−1)2
