Let's tackle the given equation step-by-step to find the missing term.
The equation is:
[tex]\[
\frac{m-n}{m^2-n^2} + \frac{?}{(m-1)(m-n)} = \frac{2m}{m^2-n^2}
\][/tex]
First, observe that [tex]\( m^2 - n^2 \)[/tex] can be factored using the difference of squares formula:
[tex]\[
m^2 - n^2 = (m - n)(m + n)
\][/tex]
With this factoring, we can rewrite the given equation:
[tex]\[
\frac{m-n}{(m-n)(m+n)} + \frac{?}{(m-1)(m-n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
Now, simplify the first term on the left-hand side:
[tex]\[
\frac{m-n}{(m-n)(m+n)} = \frac{1}{m+n}
\][/tex]
So the equation becomes:
[tex]\[
\frac{1}{m+n} + \frac{?}{(m-1)(m-n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
Next, we want to have a common denominator on both sides of the equation to combine the terms:
The common denominator between [tex]\(\frac{1}{m+n}\)[/tex] and [tex]\(\frac{2m}{(m-n)(m+n)}\)[/tex] is [tex]\((m - 1)(m - n)\)[/tex]. Multiply through by [tex]\((m-1)(m+n)(m-n)\)[/tex]:
[tex]\[
(m-1)(m+n)(m-n) \cdot \frac{1}{m+n} + (m-1)(m+n)(m-n) \cdot \frac{?}{(m-1)(m-n)} = (m-1)(m+n)(m-n) \cdot \frac{2m}{(m-n)(m+n)}
\][/tex]
Simplify each term individually:
The left-hand side's first term:
[tex]\[
(m-1)(m+n)(m-n) \cdot \frac{1}{m+n} = (m-1)(m-n)
\][/tex]
The left-hand side's second term:
[tex]\[
(m-1)(m+n)(m-n) \cdot \frac{?}{(m-1)(m-n)} = ? (m+n)
\][/tex]
And the right-hand side:
[tex]\[
(m-1)(m+n)(m-n) \cdot \frac{2m}{(m-n)(m+n)} = 2m (m-1)
\][/tex]
Combine everything:
[tex]\[
(m-1)(m-n) + ? (m+n) = 2m (m-1)
\][/tex]
Distribute and simplify:
[tex]\[
(m-1)(m-n) + ? (m+n) = 2m (m-1)
\][/tex]
[tex]\[
m^2 - mn - m + n + ? (m+n) = 2m^2 - 2m
\][/tex]
Now we isolate the term involving (?) by balancing the equation:
[tex]\[
m^2 - mn - m + n + x(m+n) = 2m^2 - 2m
\][/tex]
[tex]\[
x(m+n) = 2m^2 - 2m - m^2 + mn + m - n
\][/tex]
[tex]\[
x(m+n) = m^2 + mn - n
\][/tex]
Removing other terms from the right-hand side:
Finally:
[tex]\[
x = m - 1
\][/tex]
Thus, the missing term is:
[tex]\[
\boxed{m - 1}
\][/tex]
