Question

Is this correct?
3/x + 3/x+1 + 3/x+2.
I got the CD x(x+1)(x+2) and eventually I got (3x+3)(3x+2)/x(x+1) + (3x^2+6)/x(x+1)(x+2) + (3x^2+3)/x(x+1)(x+2)

Answer 1

This isn't my final answer i just want to know before i go on.
\[\frac{ (3x+3)(3x+6) }{ x(x+1)(x+2) } + \frac{ 3x ^{2}+6 }{ x(x+1)(x+2) } + \frac{ 3x ^{2}+3 }{ x(x+1)(x+2) }\]

Answer 2

its wrong,the first numerator should be 3(x+1)(x+2)=(3x+3)(x+2)
the second 3x(x+2)=3x^2+6x
and the third 3x(x+1)=3x^2+3x

Answer 3

OK thanks :)

Answer 4

So do I just combine the numerators next?

Answer 5

yes and see if the resulting numerator is factorizable

Answer 6

Do i multiply the (3x+3)and (x+2)?

Answer 7

yes

Answer 8

I got (x+1) (x+1)

Answer 9

\[\frac{ (x+1) }{ x(x+2) }\] is this the final answer?

Answer 10

3/(x) + 3/(x+1) + 3/(x+2)
3(x+1)(x+2)+3x(x+2)+3x(x+1)
=------------------------------------
x(x+1)(x+2)

(3x+3)(x+2)+3x^2+6x+3x^2+3x
=----------------------------------
x(x+1)(x+2)

12x^2 +18x+6
=--------------------------------
x(x+1)(x+2)
6(2x^2+3x+1)
=-------------------------------
x(x+1)(x+2)

6(x+1)(2x+1)
=----------------------
x(x+1)(x+2)
6(2x+1)
= ----------------
x(x+2)