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

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

Ответ:

nihadsalim10

29.01.2022

Mathematics

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,

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