= Solution
Use the <Charnes-Cooper transformation> on the positive-denominator region:
$$
t=\frac1{d^Tx+\beta},\qquad y=tx.
$$
It gives $Ay\le bt$, $d^Ty+\beta t=1$ and $c^Ty+\alpha t=(c^Tx+\alpha)/(d^Tx+\beta)$.
We must check that allowing $t=0$ introduces no false feasible solution. If $t=0$, then $Ay\le0$. The <recession cone> of the nonempty <linear polyhedron> $P=\{x:Ax\le b\}$ is $\{y:Ay\le0\}$. Indeed for $x_0\in P$, $x_0+uy\in P$ for every $u\ge0$. Boundedness of $P$ therefore forces $y=0$, contradicting $d^Ty=1$. Thus every transformed feasible point has $t>0$.
Conversely set $x=y/t$. The constraints give $x\in P$ and $d^Tx+\beta=1/t>0$, with the same objective value. The assumed original optimizer with positive denominator supplies a transformed feasible point. Every transformed point corresponds to an original point and has objective at most that optimizer's value. Hence \b[the transformed <linear program> attains the original optimum, and its optimizer recovers $x=y/t$.] Positivity on every point of $P$ is not needed here; the given positive-denominator optimum suffices.
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