Originally Posted by

**Sooda** This is more difficult, I see familiar things (formulas basically behind numbers), but end with different result. How it is good to start simplifying it?

Expression:

$\displaystyle \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}$

Answer:

$\displaystyle a$

Things what I see:

Formula: $\displaystyle a^2-b^2=(a+b)(a-b)$

$\displaystyle 4a^2-1=(2a+1)(2a-1)$

same thing with: $\displaystyle (a^2-1)(a-1)$

$\displaystyle (a^2-1)(a-1)=(a+1)(a-1)(a-1)$

I am not sure about this one:

$\displaystyle (a+1)(a-1)(a-1)=(a+1)(a-1)^2$

Going inside parenthesis:

Formula: $\displaystyle (a-b)^2$

$\displaystyle a^2-2a+1=(a-1)^2$

Common denominator in my opinion is:

$\displaystyle \frac{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}{(a-1)^2(a^2-1)}$

Ready to divide (which is multiplication with opposite number):

$\displaystyle \frac{(2a+1)(2a-1)*(a-1)^2(a^2-1)}{(a+1)(a-1)^2*a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}$

I know it looks crazy.

$\displaystyle \frac{(2a+1)(2a-1)*(a-1)}{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}$

I calculated it and got as final result:

$\displaystyle \frac{1}{2a}$

Interesting, but when I set a = 1 my simplification and textbook answer give different results. I went wrong, tips how I should approach it?