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    How would you simplify: \dfrac {\frac{2}{x^3}-\frac{1}{2x^2}}{\sqrt[4]{\frac{1}{x}-\frac{2}{x^3}}}?
    • PS Helper

    PS Helper
    It really depends what you mean by "simplify"... there's only so much you can do with that.

    Hey Farhan ,

    Well thats sort of a difficult question to answer because to behonest there is no significant simplification of this really, your still have your 4th root and 3rd degree x's. But one thing to make it simpler would be to remove the 'embeded' fractions withing the numerator and denominator. consider for example:

    \displaystyle \frac{2}{x^3}-\frac{1}{2x^2} = \frac{4 - x}{2x^3}

    How could this and other similar opertations to your equation be used?

    nuodai sort of beat me to it, but completly right, theres only so much you can do with this
    • Community Assistant
    • Study Helper

    Community Assistant
    Study Helper
    (Original post by Farhan.Hanif93)
    How would you simplify: \dfrac {\frac{2}{x^3}-\frac{1}{2x^2}}{\sqrt[4]{\frac{1}{x}-\frac{2}{x^3}}}?
    You can't really simplify but multiplying top and bottom by 2x^3 and grouping the fractions in the square root tidies it up a bit:

    \displaystyle \dfrac {\frac{2}{x^3}-\frac{1}{2x^2}}{\sqrt[4]{\frac{1}{x}-\frac{2}{x^3}}} = \frac{4-x}{2x^3\sqrt[4]{\frac{x^2-2}{x^3}}}

    EDIT: Way too slow.
    • Study Helper

    Study Helper
    (Original post by Farhan.Hanif93)
    Once you've done as suggested above, you can also move one of the "x"s in the denominator into the 4th root where it becomes x^4, and eliminate the fractional part under the 4th root, thus:

    \displaystyle= \frac{4-x}{2x^2\sqrt[4]{x^3-2x}}
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