1. ## Simplifying complicated fraction.

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$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{x+\frac{2}{x}}} and \displaystyle \frac{3x^2+2x+6}{x^2+2x+2} are equal when x = 2 and x = -1. True or False?/ They are equivalent expressions on (-infinity,infinity) I'm lost. How do you do this? Thank you! 2. Originally Posted by Anemori show \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{x+\frac{2}{x}2}}$ and $\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$ are equal when x = 2 and x = -1.

True or False?/ They are equivalent expressions on (-infinity,infinity)

I'm lost. How do you do this?

Thank you!
Start simplifying the left hand equation.
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}} \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2+2}{x}}}$
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2x}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2}{x^2+2} + \frac{2x}{x^2+2}}$
$\displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2x+2}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2(x^2+2)}{x^2+2x+2}$
$\displaystyle \frac{x^2+2x+2}{x^2+2x+2}+\frac{2x^2+4}{x^2+2x+2}$
$\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$

3. edit: your post is gone now i look silly :P

4. It looks false . on the fraction i will evaluate the bottom part first going up?

5. Originally Posted by Anemori
It looks false . on the fraction i will evaluate the bottom part first going up?
No, it's true. The math I showed was for when you had that extra two multiplied in. I have corrected my work to reflect your correction in the problem given.

6. Originally Posted by 1005
No, it's true. The math I showed was for when you had that extra two multiplied in. I have corrected my work to reflect your correction in the problem given.
let me see it

7. Originally Posted by Anemori
let me see it
Originally Posted by 1005
Start simplifying the left hand equation.
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}} \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2+2}{x}}}$
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2x}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2}{x^2+2} + \frac{2x}{x^2+2}}$
$\displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2x+2}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2(x^2+2)}{x^2+2x+2}$
$\displaystyle \frac{x^2+2x+2}{x^2+2x+2}+\frac{2x^2+4}{x^2+2x+2}$
$\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$

After simplifying the equation of the left, you see that they are exactly the same equation.

8. $\displaystyle 1+\frac {2}{1+\frac{2}{x+\frac{2}{x}}}$

how did you do it from that to this:

$\displaystyle 1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}$

9. Originally Posted by Anemori
$\displaystyle 1+\frac {2}{1+\frac{2}{x+\frac{2}{x}}}$

how did you do it from that to this:

$\displaystyle 1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}$
I multiplied by $\displaystyle \frac{x}{x}$ which simplifies to 1. You can do that, because 1 X anything = that same anything. I did this so the two fractions at the bottom would have the same denominator, allowing them to be added. If you continue to look at what I did, I used that method about five times to add the fractions to 1.