Question

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

Answer 1

The answer is "
`\bold{((m-1)(m-2))/((m-n))}`"

Explanation:

Given:

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

let, ? = x then,

`\Rightarrow ((m-n))/(m^2-n^2) + (x)/((m-1)(m-2)) - (2m)/(m^2-n^2)=0\\\\\Rightarrow ((m-n))/(m^2-n^2) - (2m)/(m^2-n^2)=- (x)/((m-1)(m-2)) \\\\\Rightarrow ((m-n)-2m)/((m^2-n^2)) =- (x)/((m-1)(m-2)) \\\\\Rightarrow (m-n-2m)/((m^2-n^2)) =- (x)/((m-1)(m-2)) \\\\\Rightarrow (-n-m)/((m^2-n^2)) =- (x)/((m-1)(m-2)) \\\\\Rightarrow (-(m+n))/((m+n)(m-n)) =- (x)/((m-1)(m-2)) \\\\\Rightarrow (-1)/((m-n)) =- (x)/((m-1)(m-2)) \\\\`

`\Rightarrow -((m-1)(m-2))=-x(m-n) \\\\\Rightarrow x= (- (m-1)(m-2))/(- (m-n)) \\\\\Rightarrow \boxed{x= ((m-1)(m-2))/((m-n))} \\`