Question

Question #9daa9

Answer 1

`color(green)(1/(a^2b+1))`

Explanation:

If we temporarily replace `a^2b` with `x`

`(a^2b-1)/(a^4b^2-1) rarr (x-1)/(x^2-1)`

We know that `x^2-1` can be factored as `(x-1)(x+1)`

So `(x-1)/(x^2-1)= cancel(x-1)/((cancel(x-1))(x+1))=1/(x+1)`

Restoring `x` back to `a^2b`, we have
`color(white)("XXX")(a^2b-1)/(a^4b^2-1)= 1/(a^2b+1)`