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Math Help - Simplify an Expression with Negative Exponents

  1. #1
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    Question Simplify an Expression with Negative Exponents

    please help me simplify the following expression --

    (ab)^-1 / (a^-2 + b^-2)

    if you could explain it with an appropriate property, that would be great!!

    thanks..!!
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  2. #2
    A riddle wrapped in an enigma
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    \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2  }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr  ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2}  {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}

    I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
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    Quote Originally Posted by masters View Post
    \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2  }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr  ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2}  {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}

    I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
    Looks good to me, but don't forget to specify that a \neq 0 and b \neq 0 independently.

    -Dan
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  4. #4
    Moo
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    Hi !

    A slightly different approach

    \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{1}{(ab)(a^{-2}+b^{-2})}

    -----------------------
    Using the power rule : a^b a^c=a^{b+c}

    --> (ab)a^{-2}=ba^{-1}
    -----------------------

    =\frac{1}{ba^{-1}+ab^{-1}}=\frac{1}{\frac ba+\frac ab}=\frac{1}{\frac{b^2+a^2}{ab}}=\frac{ab}{a^2+b^2  }


    one extra equality
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  5. #5
    A riddle wrapped in an enigma
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    Always keep it interesting, Moo. Good job!!
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  6. #6
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    thank you both for your help!
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  7. #7
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    Hello, needymathperson!

    One different step . . .


    \frac{(ab)^{-1}}{a^{-2} + b^{-2}}
    We have: . \frac{(ab)^{-1}} {a^{-2} + b^{-2}} \;\;=\;\;\frac{\dfrac{1}{ab}} {\dfrac{1}{a^2} + \dfrac{1}{b^2}}
    Multiply top and bottom by a^2b^2\!:\;\;\frac{a^2b^2\left(\dfrac{1}{ab}\right  )}{a^2b^2\left(\dfrac{1}{a^2} + \dfrac{1}{b^2}\right)} \;\;=\;\;\boxed{\frac{ab}{b^2+a^2}}

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