What is expand ((m-n)^2+6(m-n))/9 ?

The solution to expand ((m-n)^2+6(m-n))/9 is (m^2)/9-(2mn)/9+(n^2)/9+(2m)/3-(2n)/3

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expand ((m-n)^2+6(m-n))/9

expand

\frac{( m - n ) ^{2} + 6 ( m - n )}{9}

\frac{m ^{2}}{9} - \frac{2 mn}{9} + \frac{n ^{2}}{9} + \frac{2 m}{3} - \frac{2 n}{3}

Solution steps

\frac{( m - n ) ^{2} + 6 ( m - n )}{9}

Expand

(m - n)^{2} + 6 (m - n) : m^{2} - 2 mn + n^{2} + 6 m - 6 n

= \frac{m ^{2} - 2 mn + n ^{2} + 6 m - 6 n}{9}

Apply the fraction rule:

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

\frac{m ^{2} - 2 mn + n ^{2} + 6 m - 6 n}{9} = \frac{m ^{2}}{9} - \frac{2 mn}{9} + \frac{n ^{2}}{9} + \frac{6 m}{9} - \frac{6 n}{9}

= \frac{m ^{2}}{9} - \frac{2 mn}{9} + \frac{n ^{2}}{9} + \frac{6 m}{9} - \frac{6 n}{9}

Cancel

\frac{6 m}{9} : \frac{2 m}{3}

= \frac{m ^{2}}{9} - \frac{2 mn}{9} + \frac{n ^{2}}{9} + \frac{2 m}{3} - \frac{6 n}{9}

Cancel

\frac{6 n}{9} : \frac{2 n}{3}

= \frac{m ^{2}}{9} - \frac{2 mn}{9} + \frac{n ^{2}}{9} + \frac{2 m}{3} - \frac{2 n}{3}

\frac{2}{s ^{3} ( s + 1 )}

\frac{e ^{- m}}{( 1 + e ^{- m} ) ^{2}}

(1, - 1) y (1, 5)

e^{- 2 l n (\frac{1}{5})}

(3 x^{4} y^{7}) (5 x^{2} y)

What is expand ((m-n)^2+6(m-n))/9 ?
The solution to expand ((m-n)^2+6(m-n))/9 is (m^2)/9-(2mn)/9+(n^2)/9+(2m)/3-(2n)/3