$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
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
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
Last edited by Albert; September 7th, 2017 at 05:25.