What is (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3)) ?

The solution to (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3)) is ((k+1)(k+4))/(4(k+2)(k+3))

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simplify (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3))

simplify

\frac{k ( k + 3 )}{4 ( k + 1 ) ( k + 2 )} + \frac{1}{( k + 1 ) ( k + 2 ) ( k + 3 )}

\frac{( k + 1 ) ( k + 4 )}{4 ( k + 2 ) ( k + 3 )}

Solution steps

\frac{k ( k + 3 )}{4 ( k + 1 ) ( k + 2 )} + \frac{1}{( k + 1 ) ( k + 2 ) ( k + 3 )}

Least Common Multiplier of

4 (k + 1) (k + 2), (k + 1) (k + 2) (k + 3) : 4 (k + 1) (k + 2) (k + 3)

Adjust Fractions based on the LCM

= \frac{k ( k + 3 ) ^{2}}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )} + \frac{4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )}

Apply the fraction rule:

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

= \frac{k ( k + 3 ) ^{2} + 4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )}

Simplify

k (k + 3)^{2} + 4 : k^{3} + 6 k^{2} + 9 k + 4

= \frac{k ^{3} + 6 k ^{2} + 9 k + 4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )}

Cancel

\frac{k ^{3} + 6 k ^{2} + 9 k + 4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )} : \frac{( k + 1 ) ( k + 4 )}{4 ( k + 2 ) ( k + 3 )}

= \frac{( k + 1 ) ( k + 4 )}{4 ( k + 2 ) ( k + 3 )}

6 x^{2} + 12 x - 90

x - \frac{x}{5} = 30

8 ∣ 5 p + 8 ∣ - 5 = 11

x - x^{2}

f = 9 - (2)^{2}

What is (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3)) ?
The solution to (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3)) is ((k+1)(k+4))/(4(k+2)(k+3))