1. Simplify?

Okay first post here not sure how to do this...but I've been stuck on a two probelms...I just scanned them and am posting them as a picture...

first one is..

They both are asking to be simplified...

1)

2)

Thank's for the help!

2. start by canceling the 2's, then ...

$\frac{1}{x + \sqrt{x^2+1}} \left( 1 + \frac{x}{\sqrt{x^2+1}}\right)$

$\frac{1}{x + \sqrt{x^2+1}} \left( \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} + \frac{x}{\sqrt{x^2+1}}\right)
$

$\frac{1}{x + \sqrt{x^2+1}} \left(\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}\right)
$

can you finish?

multiply the single terms by the (-1)'s ...

$\frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$

$\frac{[(e^x + e^{-x}) - (e^x - e^{-x})] \cdot [(e^x + e^{-x}) + (e^x - e^{-x})]}{(e^x + e^{-x})^2}$

$\frac{2e^{-x} \cdot 2e^x}{(e^x + e^{-x})^2}$

$\frac{4}{(e^x + e^{-x})^2}$

3. Ok let's see if I can type this out correctly.

#1) $\frac{1}{x+\sqrt{x^2+1}} \left(1 + \frac{2x}{2 \sqrt{x^2+1}} \right)$

So distribute the first expression through the parenthesis.

$\frac{1}{x+\sqrt{x^2+1}}+\left( \frac{1}{x+\sqrt{x^2+1}} \right) \left(\frac{2x}{2 \sqrt{x^2+1}} \right)$

Multiply the last two expressions.

$\frac{1}{x+\sqrt{x^2+1}}+\frac{2x}{2x \sqrt{x^2+1}+2(x^2+1)}$

The denominator of expression two can be factored with some moving around.

$\frac{1}{x+\sqrt{x^2+1}}+\frac{2x}{(x+\sqrt{x^2+1} )^2-1}$

Now multiply each expression by the conjugate of the denominator to get the radicals gone.

EDIT: I'll keep the post here just because it took me so long, but look above at skeeter's post for a much quicker, better simplification.

4. thanks skeeter but I don't understand the 2nd step for the first question....also i don't know what to do after the last step you did...

I think i get the 2nd question...is that the final answer?

But thanks for the help guys!

5. Skeeter found the common denominator for the second step

For the last step cancel out

$x + \sqrt{x^2+1}$

6. Originally Posted by 11rdc11
Skeeter found the common denominator for the second step

For the last step cancel out

$x + \sqrt{x^2+1}$
how do i cancel out $x + \sqrt{x^2+1}$ because they appear twice on one side and once on the other...

thanks mate

7. $\frac{1}{x + \sqrt{x^2+1}} \left(\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}\right)$

Ok so you can see this easier lets

$(x + \sqrt{x^2+1}) = a$

so it would look like this

$\bigg(\frac{1}{a}\bigg) \bigg(\frac{a}{\sqrt{x^2+1}}\bigg)$

8. so the final answer is 1/root x^2 + 1?

9. actually nvm it would be root x^2+1

10. $\frac{1}{\sqrt{x^2+1}}$