Sure, let’s find the missing term that makes the equation true. The equation given is:
[tex]\[
\frac{m-n}{m^2 - n^2} + \frac{\square}{(m-1)(m-n)} = \frac{2m}{m^2 - n^2}
\][/tex]
### Step-by-Step Solution:
1. Factor the denominator [tex]\(m^2 - n^2\)[/tex]:
[tex]\[
m^2 - n^2 = (m-n)(m+n)
\][/tex]
So, the equation can be rewritten as:
[tex]\[
\frac{m-n}{(m-n)(m+n)} + \frac{\square}{(m-1)(m-n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
2. Simplify the first term:
[tex]\[
\frac{m-n}{(m-n)(m+n)} = \frac{1}{m+n}
\][/tex]
So, now the equation is:
[tex]\[
\frac{1}{m+n} + \frac{\square}{(m-1)(m-n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
3. Find a common denominator for the left side:
The common denominator for the left side terms is [tex]\((m-1)(m-n)(m+n)\)[/tex]. Rewriting our equation with this common denominator:
[tex]\[
\frac{(m-1)(m-n)}{(m+n)(m-1)(m-n)} + \frac{\square (m+n)}{(m-1)(m-n)(m+n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
4. Combine the fractions on the left side:
[tex]\[
\frac{m-1}{(m-1)(m-n)(m+n)} + \frac{\square}{(m-1)(m-n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
Since the denominators are now the same, we can combine the numerators:
[tex]\[
\frac{(m-1) + \square}{(m-1)(m-n)(m+n)} = \frac{2m}{(m-n)(m+n)}
\][/tex]
5. Equate the numerators:
By comparing the numerators of the fractions:
[tex]\[
(m-1) + \square = 2m
\][/tex]
6. Solve for [tex]\(\square\)[/tex]:
[tex]\[
\square = 2m - (m-1)
\][/tex]
Simplify the right-hand side:
[tex]\[
\square = 2m - m + 1 = m + 1
\][/tex]
### Conclusion:
The missing term is [tex]\(\boxed{m + 1}\)[/tex].
Therefore, to make the equation true, you should replace the question mark with:
[tex]\[
\boxed{m + 1}
\][/tex]
