Simplifying complicated fraction.

**Anemori** · January 31st 2010, 07:28 AM

show
$1+$ $\frac {2}{1+\frac{2}{x+\frac{2}{x}}}$ and $\frac{3x^2+2x+6}{x^2+2x+2}$ 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!

**1005** · January 31st 2010, 08:09 AM

Originally Posted by Anemori

show
$1+$ $\frac {2}{1+\frac{2}{x+\frac{2}{x}2}}$ and $\frac{3x^2+2x+6}{x^2+2x+2}$ 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.
$1+$ $\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}$
$1+$ $\frac {2}{1+\frac{2}{\frac{x^2+2}{x}}}$
$1+$ $\frac {2}{1+\frac{2x}{x^2+2}}$
$1+$ $\frac{2}{\frac{x^2+2}{x^2+2} + \frac{2x}{x^2+2}}$
$1+$ $\frac{2}{\frac{x^2+2x+2}{x^2+2}}$
$1+$ $\frac{2(x^2+2)}{x^2+2x+2}$
$\frac{x^2+2x+2}{x^2+2x+2}+\frac{2x^2+4}{x^2+2x+2}$
$\frac{3x^2+2x+6}{x^2+2x+2}$

**1005** · January 31st 2010, 08:14 AM

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

**Anemori** · January 31st 2010, 08:20 AM

It looks false

. on the fraction i will evaluate the bottom part first going up?

**1005** · January 31st 2010, 08:25 AM

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.

**Anemori** · January 31st 2010, 08:54 AM

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

**1005** · January 31st 2010, 09:51 AM

Originally Posted by Anemori

let me see it

Originally Posted by 1005

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

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

**Anemori** · January 31st 2010, 10:46 AM

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

how did you do it from that to this:

$1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}<br />$

**1005** · January 31st 2010, 01:58 PM

Originally Posted by Anemori

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

how did you do it from that to this:

$1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}<br />$

I multiplied by $\frac{x}{x}$ 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.

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