What is 1/(m-n)+m/(n^2-m^2) ?

The solution to 1/(m-n)+m/(n^2-m^2) is n/((m-n)(m+n))

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simplify 1/(m-n)+m/(n^2-m^2)

simplify

\frac{1}{m - n} + \frac{m}{n ^{2} - m ^{2}}

\frac{n}{( m - n ) ( m + n )}

Solution steps

\frac{1}{m - n} + \frac{m}{n ^{2} - m ^{2}}

Least Common Multiplier of

m - n, n^{2} - m^{2} : (m - n) (m + n) (n - m)

Adjust Fractions based on the LCM

= \frac{- ( m - n ) ( m + n )}{( m - n ) ( m + n ) ( n - m )} + \frac{m ( m - n )}{( m - n ) ( m + n ) ( n - m )}

Apply the fraction rule:

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

= \frac{- ( m - n ) ( m + n ) + m ( m - n )}{( m - n ) ( m + n ) ( n - m )}

Simplify

- (m - n) (m + n) + m (m - n) : n^{2} - mn

= \frac{n ^{2} - mn}{( m - n ) ( m + n ) ( n - m )}

Cancel

\frac{n ^{2} - mn}{( m - n ) ( m + n ) ( n - m )} : \frac{n}{( m - n ) ( m + n )}

= \frac{n}{( m - n ) ( m + n )}

(2 x + 5 y) + (3 x - 2 y)

\frac{49}{5}

\frac{49}{9}

(5 - 4 i) (5 + 4 i)

2^{l o g_{2} (4)}

What is 1/(m-n)+m/(n^2-m^2) ?
The solution to 1/(m-n)+m/(n^2-m^2) is n/((m-n)(m+n))