Question

closed

Answer 1

and we want to express it as a single fully simplified fraction (i'm assuming) ?

Answer 2

yes

Answer 3

\[\frac{3}{x} -\frac{3}{x+1} + \frac{3}{x+2}\]

we want a common denominator

in this case the common denominator will be x(x+1)(x+2)

as it is the lowest common multiple of the initial denominators

lets do it term by term:
\[\frac{3}{x} = \frac{3(x+1)(x+2)}{x(x+1)(x+2)}\]

\[\frac{3}{x+1} = \frac{3x(x+2)}{x(x+1)(x+2)}\]

\[\frac{3}{x+2} = \frac{3x(x+1)}{x(x+1)(x+2)}\]

Answer 4

yes thank you

Answer 5

so we have:
\[\frac{3}{x} + \frac{3}{x+1} + \frac{3}{x+2} = \frac{3(x+1)(x+2)}{x(x+1)(x+2)} + \frac{3x(x+2)}{x(x+1)(x+2)} + \frac{3x(x+1)}{x(x+1)(x+2)}\]

Answer 6

now you can add them

Answer 7

phew that typing was a mission

Answer 8

yes it was lol but thank you so much i needed this help

Answer 9

its fine :)