Question

How can (x^2)/(x-1) simplify to (x-1) + 1/(x+1) ??

Answer 1

\[\frac{ x^2 }{ x-1 } = x-1 + \frac{ 1 }{ x + 1 }\]

Answer 2

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

Answer 3

Hmm I get how it happens backwards, but what can I do to x^2/x-1 to get the subsequent answer?

Answer 4

because its not true

\(\frac{ x^2 }{ x-1 } = x-1 + \frac{ 1 }{ x + 1 }\) is only true when \(x=0\)

Answer 5

http://www.wolframalpha.com/input/?i= \frac{+x^2+}{+x-1+}+%3D+x-1+%2B+\frac{+1+}{+x+%2B+1+}

Answer 6

Sorry you were right. The real expression should look like \[\frac{ x^2 }{ x+1 } = x-1 + \frac{ 1 }{ x+1 }\]

Answer 7

Its not intuitive how to go "backwards" but its the process above, just done backwards.

Answer 8

Long division can help with the "backwards" process.

Answer 9

yeah, true. I think after a while you also just get used to the old x=x+a-a trick...

Answer 10

But if I just want to show its true, and you already know the answer. Then no reason to do any work. Just show its true. Now if you are trying to figure out how to get its "linear" parts separated out of something like this, then do long division.

Answer 11

But I would google that if you dont know how. its hell to do in latex.

Answer 12

What exactly do you do to \[\frac{ x^2 }{ x+1 }\] to get x-1 + 1/(x+1) was the part that bothered me because I don't understand the process.