Question

divide.

Answer 1

3n^2 - n/ n^2 - 1 / n^2/ n+1

Answer 2

\[\frac{ 3n^2-n }{ n^2-1}\div \frac{ n^2 }{ n+1 }\]

is that what you mean?

Answer 3

yes :)

Answer 4

Ok, recognize that n^2-1 is a perfect square, and then we can rewrite the problem as
\[\frac{ 3n^2-n }{ (n-1)(n+1) }*\frac{ n+1 }{ n^2 }\]

Answer 5

Hope that helps

Answer 6

\[\frac{ 3n ^{2}- n }{ n-1 } \times \frac{ 1 }{ n ^{2} }\]

Answer 7

shall i also cancel the n^2 out?

Answer 8

I dont think so without getting it messy. But factor out a n from the 3n^2 and cancel

Answer 9

so the answer is \[\frac{ 3-n }{ n-1}\] ?

Answer 10

Uhmm not quite,

Answer 11

\[\frac{ 3n-1}{ n^2-n }\]