+130071 1 / [ 2m - 4n ] - 1 / [ 2m + 4n ] - m / [ m^2 - 4n^2] factor the denominators
1 / [ 2 (m - 2n) ] - 1 / [ 2 (m + 2n) ] - m / [ (m + 2n) (m - 2n) ]
The common denominator is 2 (m + 2n) (m - 2n).....so we have
( [ (m + 2n) ] - [m - 2n] - [2m] ) / [ 2 (m + 2n) (m - 2n) ]
( 4n - 2m ] / [ 2 (m + 2n) (m - 2n) ]
[ 2 (2n - m) ] / [ 2 (m + 2n) (m - 2n) ]
( 2n - m) / [ (m + 2n) (m - 2n) ]
- ( m - 2n) / [ (m + 2n) (m - 2n) ]
- 1 / [ m + 2n ]
+9481 Are you asking about this part:
( [ (m + 2n) ] - [m - 2n] - [2m] ) / [ 2 (m + 2n) (m - 2n) ] ???
Starting from the previous step...
\(\frac{1}{2(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{m}{(m+2n)(m-2n)} \)
Multiply the third fraction by 2/2 .
\(=\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{2m}{2(m+2n)(m-2n)} \\~\\ =\, \frac{(m+2n)-(m-2n)-(2m)}{2(m+2n)(m-2n)}\)
