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Math Help - [SOLVED] division of an algebric expression - no 2 (and last)

  1. #1
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    Question [SOLVED] division of an algebric expression - no 2 (and last)

    Here I go one more that I'm really not sure of, what I really hate is that we have all that practice homework but none of the exercise where as complexe as those in the actual homework, so I basically have no reference...

    <br /> <br />
\left( \frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right)^{-2}<br /> <br />

    \left( x^{-1-(-4)}(x-y)^{\frac {3}{1}-\frac {1}{2}}(y^{1-3})^\frac {1}{4} \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {6}{2}-\frac {1}{2}}(y^2)^\frac {1}{4} \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {5}{2}}(y^{2^(\frac {1}{4})}) \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {5}{2}}y^{\frac {1}{2}} \right)^{-2}<br />


    x^{-6}(x-y)^{-5}y^{-1}

    Can I simplify this even further? The answer cannot contain fractions or radicals.

    Thank you so much, for all the help, I'm eternaly greatful!
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  2. #2
    Eater of Worlds
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    You probably want to express it without negative powers.

    It should simplify to \frac{y^{\frac{1}{2}}}{x^{6}(x-y)^{5}}
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  3. #3
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    Hello, Neenoon!

    Your work is correct . . . This is the way I organize the parts.


    \left[\frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right]^{-2}

    We have: . \left[\frac{x^{\text{-}1}}{x^{-4}}\cdot \frac{\sqrt[4]{y^3}}{y}\cdot\frac{(x-y)^3}{\sqrt{x-y}}\right]^{-2} . = \;\left(\frac{x^{-1}}{x^{-4}}\cdot\frac{y^{\frac{3}{4}}}{y^1}\cdot\frac{(x-y)^3}{(x-y)^{\frac{1}{2}}} \right)^{-2}

    . . = \;\left[x^{-1-(-4)} \cdot y^{\frac{3}{4}-1} \cdot(x-y)^{3-\frac{1}{2}}\right]^{-2} \;=\;\left[x^3\cdot y^{-\frac{1}{4}}\cdot(x-y)^{\frac{5}{2}}\right]^{-2}

    . . = \;\left[x^3\right]^{\text{-}2}\cdot\left[y^{-\frac{1}{4}}\right]^{-2}\cdot\left[(x-y)^{\frac{5}{2}}\right]^{-2} \;=\;x^{-6}\cdot y^{\frac{1}{2}}\cdot(x-y)^{-5}

    . . =\;\frac{\sqrt{y}}{x^6\,(x-y)^5}

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  4. #4
    Super Member 11rdc11's Avatar
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    Isn't this step wrong because the final part should be sqrt y? Also he said his answer could not contain radicals or fractions, so how would he get rid of the radical sqrt(y)?
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by 11rdc11 View Post
    Isn't this step wrong because the final part should be sqrt y? Also he said his answer could not contain radicals or fractions, so how would he get rid of the radical sqrt(y)?
    Just stop at this step then:

    \;\left[x^3\right]^{\text{-}2}\cdot\left[y^{-\frac{1}{4}}\right]^{-2}\cdot\left[(x-y)^{\frac{5}{2}}\right]^{-2} \;=\;x^{-6}\cdot y^{\frac{1}{2}}\cdot(x-y)^{-5}<br />
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  6. #6
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    Quote Originally Posted by Neenoon View Post
    Here I go one more that I'm really not sure of, what I really hate is that we have all that practice homework but none of the exercise where as complexe as those in the actual homework, so I basically have no reference...

    <br /> <br />
\left( \frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right)^{-2}<br /> <br />

    \left( x^{-1-(-4)}(x-y)^{\frac {3}{1}-\frac {1}{2}}{\color{red}(y^{\frac{3}{4}-1})} \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {6}{2}-\frac {1}{2}}{\color{red}(y^{-\frac{1}{4}})} \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {5}{2}} {\color{red}(y^{-\frac{1}{4}})} \right)^{-2}<br />

    \left( x^3(x-y)^{\frac {5}{2}}{\color{red}y^{-\frac {1}{4}} }\right)^{-2}<br />


    x^{-6}(x-y)^{-5}{\color{red}y^{\frac{1}{2}}}

    Can I simplify this even further? The answer cannot contain fractions or radicals.

    Thank you so much, for all the help, I'm eternaly greatful!
    Soroban, your work is impeccable. I have made corrections in the original poster's work. There was a slight problem with the y variable.
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  7. #7
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