Evaluate:
(3−1÷4−1)2(3^{-1} ÷ 4^{-1})^2(3−1÷4−1)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÷4−1)2=(131÷141)2=(13÷14)2=(13×41)2=(1×43×1)2=(43)2=(4×43×3)=(169)=1(79)(3^{-1} ÷ 4^{-1})^2\\[1em] = \Big(\dfrac{1}{3}^1 ÷ \dfrac{1}{4}^1\Big)^2\\[1em] = \Big(\dfrac{1}{3} ÷ \dfrac{1}{4}\Big)^2\\[1em] = \Big(\dfrac{1}{3} \times \dfrac{4}{1}\Big)^2\\[1em] = \Big(\dfrac{1 \times 4}{3 \times 1}\Big)^2\\[1em] = \Big(\dfrac{4}{3}\Big)^2\\[1em] = \Big(\dfrac{4 \times 4}{3 \times 3}\Big)\\[1em] = \Big(\dfrac{16}{9}\Big)\\[1em] = 1\Big(\dfrac{7}{9}\Big)(3−1÷4−1)2=(311÷411)2=(31÷41)2=(31×14)2=(3×11×4)2=(34)2=(3×34×4)=(916)=1(97)
Hence, (3−1÷4−1)2=1(79)(3^{-1} ÷ 4^{-1})^2 = 1\Big(\dfrac{7}{9}\Big)(3−1÷4−1)2=1(97)
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