Question

M-n/m^2-n^2 + ?/(m-1)(m-2) - 2m/m^2-n^2

Answer 1

Answer:

The answer is "$\bold{\frac{(m-1)(m-2)}{(m-n)}}$"

Step-by-step explanation:

Given:

$\bold{\frac{(m-n)}{m^2-n^2} + \frac{?}{(m-1)(m-2)} - \frac{2m}{m^2-n^2}=0}$

let,  ? = x then,

$\Rightarrow \frac{(m-n)}{m^2-n^2} + \frac{x}{(m-1)(m-2)} - \frac{2m}{m^2-n^2}=0\\\\\Rightarrow \frac{(m-n)}{m^2-n^2} - \frac{2m}{m^2-n^2}=- \frac{x}{(m-1)(m-2)} \\\\\Rightarrow \frac{(m-n)-2m}{(m^2-n^2)} =- \frac{x}{(m-1)(m-2)} \\\\\Rightarrow \frac{m-n-2m}{(m^2-n^2)} =- \frac{x}{(m-1)(m-2)} \\\\\Rightarrow \frac{-n-m}{(m^2-n^2)} =- \frac{x}{(m-1)(m-2)} \\\\\Rightarrow \frac{-(m+n)}{(m+n)(m-n)} =- \frac{x}{(m-1)(m-2)} \\\\\Rightarrow \frac{-1}{(m-n)} =- \frac{x}{(m-1)(m-2)}$

$\Rightarrow -((m-1)(m-2))=-x(m-n) \\\\\Rightarrow x= \frac{- (m-1)(m-2)}{- (m-n)} \\\\\Rightarrow \boxed{x= \frac{(m-1)(m-2)}{(m-n)}}$