Prove that: (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}(m+n)−1(m−1+n−1)=(mn)−1
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To Prove: (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}(m+n)−1(m−1+n−1)=(mn)−1
Taking LHS:
(m+n)−1(m−1+n−1)=(1m+n)(1m+1n)=(1m+n)(nmn+mnm)=(1m+n)(n+mmn)=1×(n+m)(m+n)×mn=1mn=(mn)−1=RHS(m + n)^{-1}(m^{-1} + n^{-1})\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{1}{m} + \dfrac{1}{n}\Big)\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{n}{mn} + \dfrac{m}{nm}\Big)\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{n + m}{mn}\Big)\\[1em] = \dfrac{1 \times (n + m)}{(m + n) \times mn}\\[1em] = \dfrac{1}{mn}\\[1em] = (mn)^{-1} \\[1em] = \text{RHS}(m+n)−1(m−1+n−1)=(m+n1)(m1+n1)=(m+n1)(mnn+nmm)=(m+n1)(mnn+m)=(m+n)×mn1×(n+m)=mn1=(mn)−1=RHS
∴ LHS = RHS
Hence, (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}(m+n)−1(m−1+n−1)=(mn)−1
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