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    Íá ÇáäÙÇã Ýì Ø

    \displaystyle{\begin{cases}x^{x+y}=y^n \\ y^{x+y}=x^{2n}y^n \end{cases},n \in \mathbb N^{*}}

  2. The Following 2 Users Say Thank You to Úáì ãÍãÏ ÇáãÕÑì For This Useful Post:

     (06-22-2016),  (06-22-2016)

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    ääÊÙÑ ÇáÍá ãä ÇáãÈÏÚíä

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    ÑÏ: Íá ÇáäÙÇã Ýì Ø

    $\large {\color{DarkBlue} let, x+y=m\Rightarrow x^{m}=y^{n}\Rightarrow mlnx=nlny\Rightarrow lnx=\frac{nlny}{m}}$
    ${\color{DarkBlue} y^{m}=x^{2n}y^{n}\Rightarrow mlny=2nlnx+nlny\Rightarrow mlny=\frac{2n^{2}lny}{m}+nlny}$
    ${\color{DarkBlue} \Rightarrow m^{2}-mn-2n^{2}=0\rightarrow (m-2n)(m+n)=0} $
    ${\color{DarkBlue} if,m=2n\rightarrow x^{2n}=y^{n}\rightarrow y=x^{2}\rightarrow S1={(1,1),(2,4),...}}$
    $ {\color{DarkBlue} if,m=-n\rightarrow x^{-n}=y^{n}\rightarrow y=\frac{1}{x}\rightarrow S2=(1,1)}$
    ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ Ìáíá ÇáÒíÇÏí ; 06-22-2016 ÇáÓÇÚÉ 04:54 PM

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