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Thread: Simplify

  1. #1
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    Simplify

    Hi

    I need to simplify the following, giving the result without fractional indices:

    $\displaystyle \frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}} $

    The solution: $\displaystyle (x+1)^2 \sqrt{x^2-1} $

    I don't see how they obtain this solution, can somebody give me a few steps?

    Thank you
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  2. #2
    Junior Member
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    Quote Originally Posted by Coxx View Post
    Hi

    I need to simplify the following, giving the result without fractional indices:

    $\displaystyle \frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}} $

    The solution: $\displaystyle (x+1)^2 \sqrt{x^2-1} $

    I don't see how they obtain this solution, can somebody give me a few steps?

    Thank you
    simplify:
    $\displaystyle \frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}} $
    ----------------------------------------------------------
    remember:
    $\displaystyle (a^2-b^2)=(a-b)(a+b)$
    so,
    $\displaystyle (x^2-1)^2= ((x-1)(x+1))^2=(x+1)^2 (x-1)^2$
    -----------------------------------------------------------
    $\displaystyle \frac{(x+1)^2 (x-1)^2 \sqrt{x+1}}{(x-1)^{3/2}} $

    u can cancel (x-1)...
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  3. #3
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    Lexington, MA (USA)
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    Hello, Coxx!

    What a strange way to leave the answer . . .


    Simplify: .$\displaystyle \frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}} $

    The solution: $\displaystyle (x+1)^2 \sqrt{x^2-1} $

    We have: .$\displaystyle \frac{\bigg[(x-1)(x+1)\bigg]^2\sqrt{x+1}}{(x-1)^{\frac{3}{2}}} $ .$\displaystyle = \;\frac{(x-1)^2(x+1)^2(x+1)^{\frac{1}{2}}}{(x-1)^{\frac{3}{2}}} $

    . . . . $\displaystyle =\; \frac{(x-1)^2}{(x-1)^{\frac{3}{2}}} \cdot\frac{(x+1)^2(x+1)^{\frac{1}{2}}}{1} \;=\;(x-1)^{\frac{1}{2}}(x+1)^{\frac{1}{2}}(x+1)^2$

    . . . . $\displaystyle =\;\bigg[(x-1)(x+1)\bigg]^{\frac{1}{2}}(x+1)^2 \;=\;(x^2-1)^{\frac{1}{2}}(x+1)^2 $

    . . . . $\displaystyle = \;\sqrt{x^2-1}\,(x+1)^2$

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  4. #4
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    hi,

    So obvious many thanks

    I just started to improve my maths skills so i'm really a newbie in this field!

    greets
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