1. ## Simplifying Expression 4

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

$\displaystyle a$

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: $\displaystyle (a^2-1)(a-1)$

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

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

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?

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

$\displaystyle a$

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

becomes

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

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

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

$\displaystyle \frac{4a^2-1}{(a+1)(a-1)^2}\cdot \left(\frac{(a-1)^2 (a+1)}{2(2a-1)}\right)-\frac{1}{2}$

$\displaystyle \dfrac{2a+1}2-\dfrac12=a+\dfrac12-\dfrac12=a$

3. 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:
$\displaystyle \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}$

$\displaystyle a$

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: $\displaystyle (a^2-1)(a-1)$

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

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

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?
ok
the denominator

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

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

the common denominator is $\displaystyle (1-a)^2(a+1)$ note : $\displaystyle (a-1)^2=(1-a)^2$

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

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

$\displaystyle \frac{4a-2}{(1-a)^2(a+1)}$ ok now we have

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

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

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

$\displaystyle \frac{2a+1}{2}-\frac{1}{2}=a$

4. How you know $\displaystyle a^2-2a+1$ has $\displaystyle (a-1)$ already in it?

EDIT:
$\displaystyle -\frac{a}{(1-a)(1+a)}=\frac{a}{(a-1)(a+1)}$
?

I am confused how arguments move in denominator, by what rule?

EDIT2:
I have never seen something like that:
$\displaystyle (a-1)^2=(1-a)^2$

Is it only true with $\displaystyle (a-1)^2=(1-a)^2$ or would $\displaystyle (4-3)^2=(3-4)^2$ ?

5. How you know $\displaystyle a^2-2a+1$ has $\displaystyle (a-1)$ already in it?

EDIT:
$\displaystyle -\frac{a}{(1-a)(1+a)}=\frac{a}{(a-1)(a+1)}$
?
do you know how to find roots of quartic polynomials ??

for the second question

$\displaystyle -\frac{a}{(1-a)(1+a)}=\frac{-a}{(1-a)(1+a)}=\frac{a}{-(1-a)(1+a)}$ since $\displaystyle \frac{-a}{b} = \frac{a}{-b}$

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

EDIT2:
I have never seen something like that:
$\displaystyle (a-1)^2=(1-a)^2$

Is it only true with $\displaystyle (a-1)^2=(1-a)^2$ or would $\displaystyle (4-3)^2=(3-4)^2$ ?
try to expand it you will see that it is the same if you do not want to expand it

you know that any negative with square power will be positive

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

suppose that $\displaystyle (1-a)$ positive number then $\displaystyle (a-1)$ will be negative but if you take the square power the two number will be the same

$\displaystyle -3^2=9 , 3^2=9$

is there anything not clear ??

6. I overlooked something obvious here, afterwards I ask: "How you know
$\displaystyle a^2-2a+1$ has $\displaystyle (a-1)$ in it? and I wrote in my first post:
Going inside parenthesis:

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

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

> do you know how to find roots of quartic polynomials ??