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Math Help - Simplify?

  1. #1
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    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!
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  2. #2
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    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)<br />

    \frac{1}{x + \sqrt{x^2+1}} \left(\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}\right)<br />

    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}
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  3. #3
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    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.
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  4. #4
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    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!
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  5. #5
    Super Member 11rdc11's Avatar
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    Skeeter found the common denominator for the second step

    For the last step cancel out

    x + \sqrt{x^2+1}
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  6. #6
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    Quote Originally Posted by 11rdc11 View Post
    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
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  7. #7
    Super Member 11rdc11's Avatar
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    \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)
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  8. #8
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    so the final answer is 1/root x^2 + 1?
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  9. #9
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    actually nvm it would be root x^2+1
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  10. #10
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    \frac{1}{\sqrt{x^2+1}}
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