Simplify the following:
a−1+b−1(ab)−1\dfrac{a^{-1} + b^{-1}}{(ab)^{-1}}(ab)−1a−1+b−1
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Given,
⇒a−1+b−1(ab)−1=(a−1+b−1)×1(ab)−1=(1a+1b)×ab=(a+bab)×ab=a+b.\Rightarrow \dfrac{a^{-1} + b^{-1}}{(ab)^{-1}} = (a^{-1} + b^{-1}) \times \dfrac{1}{(ab)^{-1}} \\[1em] = \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \times ab \\[1em] = \Big(\dfrac{a + b}{ab}\Big) \times ab \\[1em] = a + b.⇒(ab)−1a−1+b−1=(a−1+b−1)×(ab)−11=(a1+b1)×ab=(aba+b)×ab=a+b.
Hence, a−1+b−1(ab)−1\dfrac{a^{-1} + b^{-1}}{(ab)^{-1}}(ab)−1a−1+b−1 = a + b.
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