If m ≠ n and (m + n)^-1 (m^-1 + n^-1) = m^xn^y;

show that : x + y + 2 = 0

Given,

$\Rightarrow (m + n)^{-1}(m^{-1} + n^{-1}) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{1}{m} + \dfrac{1}{n}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{n + m}{mn}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{mn} = m^xn^y \\[1em] \Rightarrow m^{-1}n^{-1} = m^xn^y \\[1em] \Rightarrow x = -1 \text{ and } y = -1.$

Substituting value in L.H.S. of equation x + y + 2 = 0, we get :

⇒ x + y + 2 = (-1) + (-1) + 2 = -2 + 2 = 0.

Since, L.H.S. = R.H.S. = 0.

Hence, proved that x + y + 2 = 0.