What is ((mn)/5-1)(m^2n^2+4mn) ?

The solution to ((mn)/5-1)(m^2n^2+4mn) is ((-5+mn)(m^2n^2+4mn))/5

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simplify ((mn)/5-1)(m^2n^2+4mn)

simplify

(\frac{mn}{5} - 1) (m^{2} n^{2} + 4 mn)

\frac{( - 5 + mn ) ( m ^{2} n ^{2} + 4 mn )}{5}

Solution steps

(\frac{mn}{5} - 1) (m^{2} n^{2} + 4 mn)

Join

\frac{mn}{5} - 1 : \frac{- 5 + mn}{5}

= \frac{- 5 + mn}{5} (m^{2} n^{2} + 4 mn)

Apply the fraction rule:

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

= \frac{( - 5 + mn ) ( m ^{2} n ^{2} + 4 mn )}{5}

3 (8 - \frac{c ^{2}}{2})

(1 + \frac{1}{n} i)^{2}

(x^{3} y + x^{3} y^{2}) d x + x^{2} y d y

(- 36 sin^{3} (r) cos (t) + 6 cos^{2} (t)) d t

\frac{s}{( s - 10 ) ( s ^{2} + 5 )}

What is ((mn)/5-1)(m^2n^2+4mn) ?
The solution to ((mn)/5-1)(m^2n^2+4mn) is ((-5+mn)(m^2n^2+4mn))/5