What is expand ((6A+2k+1)(k+1))/k ?

The solution to expand ((6A+2k+1)(k+1))/k is 6A+(6A)/k+2k+3+1/k

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expand ((6A+2k+1)(k+1))/k

expand

\frac{( 6 A + 2 k + 1 ) ( k + 1 )}{k}

6 A + \frac{6 A}{k} + 2 k + 3 + \frac{1}{k}

Solution steps

\frac{( 6 A + 2 k + 1 ) ( k + 1 )}{k}

Expand

(6 A + 2 k + 1) (k + 1) : 6 A k + 6 A + 2 k^{2} + 3 k + 1

= \frac{6 A k + 6 A + 2 k ^{2} + 3 k + 1}{k}

Apply the fraction rule:

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

\frac{6 A k + 6 A + 2 k ^{2} + 3 k + 1}{k} = \frac{6 A k}{k} + \frac{6 A}{k} + \frac{2 k ^{2}}{k} + \frac{3 k}{k} + \frac{1}{k}

= \frac{6 A k}{k} + \frac{6 A}{k} + \frac{2 k ^{2}}{k} + \frac{3 k}{k} + \frac{1}{k}

Cancel

\frac{6 A k}{k} : 6 A

= 6 A + \frac{6 A}{k} + \frac{2 k ^{2}}{k} + \frac{3 k}{k} + \frac{1}{k}

Cancel

\frac{2 k ^{2}}{k} : 2 k

= 6 A + \frac{6 A}{k} + 2 k + \frac{3 k}{k} + \frac{1}{k}

Cancel

\frac{3 k}{k} : 3

= 6 A + \frac{6 A}{k} + 2 k + 3 + \frac{1}{k}

(4 x + 4 δ x) \cdot (x^{2} - 4)

\frac{24}{( s - 2 ) ^{4}}

2 (- 3 x^{2})

\frac{4 x ^{2} - 8 x + 4}{( x + 1 ) ^{2}}

\frac{( - 2 y + 2 ) ^{3}}{3}

What is expand ((6A+2k+1)(k+1))/k ?
The solution to expand ((6A+2k+1)(k+1))/k is 6A+(6A)/k+2k+3+1/k