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Let the numerator of the fraction be $x$, then its denominator is $2x+1$.

So the fraction is $2x+1xβ$

Then $2x+1xβ+x2x+1β=22116β=2158β$

$β2x_{2}+xx_{2}+4x_{2}+4x+1β=2158β$

$β2x_{2}+x5x_{2}+4x+1β=2158β$

$β105x_{2}+84x+21=116x_{2}+58x$

$β11x_{2}β26xβ21=0$

$β11x_{2}β33x+7xβ21=0$

$β11x(xβ3)+7(xβ3)=0$

$β11x=β7$ or $x=3$

$β΄x=11β7β,3$

$βx=3$ since $x$ is a natural number $β΄xβ₯0$

$β΄$ Required fraction is $2x+1xβ=2Γ3+13β=73β$

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