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    #1
    $(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
    $find$: $x,y$
    $(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
    $find$: $x,y$
    $(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
    $find$: $x,y,z$

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    #2
    Quote Originally Posted by Albert View Post
    $(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
    $find$: $x,y$
    $(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
    $find$: $x,y$
    $(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
    $find$: $x,y,z$
    without loss of generality we can take x <=y <=z as any permutation can give ans

    for (1)



    for (2)

    Last edited by kaliprasad; September 5th, 2017 at 11:22. Reason: corrected the value based on albert;s comment

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    Quote Originally Posted by kaliprasad View Post
    without loss of generality we can take x <=y <=z as any permutation can give ans

    for (1)



    for (2)


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    #4

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    Quote Originally Posted by Albert View Post
    $(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
    $find$: $x,y$
    $(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
    $find$: $x,y$
    $(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
    $find$: $x,y,z$
    hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

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    #6
    Quote Originally Posted by Albert View Post
    hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
    The above hint is not correct because


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    #7
    Quote Originally Posted by Albert View Post
    hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

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    Last edited by Albert; September 7th, 2017 at 05:25.

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    #9
    Quote Originally Posted by Albert View Post
    any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me
    These are known as .

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    Quote Originally Posted by Opalg View Post

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