= Solution
Let $D=Q/V_0$ be the <dilution rate>. Population loss is $DN$, and nutrient loss by consumption is $NK(C)/V_0$, so <mass conservation> gives
$$
\dot N=\gamma NK(C)-DN,\qquad \dot C=D(C_0-C)-\frac{NK(C)}{V_0}.
$$
Use the dimensionless variables $\tau=Dt$, $c=C/K_0$, and $n=NK_{\max}/(QK_0)$. Substitution into the <Monod equation> gives the stated system with
$$
\boxed{\alpha=\frac{\gamma K_{\max}V_0}{Q},\qquad\beta=\frac{C_0}{K_0}.}
$$
For a positive <equilibrium>, the first equation requires $\alpha c/(1+c)=1$. Hence $c_*=1/(\alpha-1)$ and $n_*=\alpha(\beta-c_*)$, both positive exactly under the specified inequalities.
Writing $A=\alpha n_* /(1+c_*)^2>0$, the <Jacobian matrix> at this <equilibrium> is
$$
J=\begin{pmatrix}0&A\\-1/\alpha&-1-A/\alpha\end{pmatrix},\qquad
\boxed{\operatorname{spec}J=\{-1,-A/\alpha\}.}
$$
Both <eigenvalues> are negative, so the state is <locally asymptotically stable>. The <Monod chemostat equilibrium and relaxation> also makes one decay rate transparent: $z=c+n/\alpha$ satisfies $z'=\beta-z$, independently of the uptake kinetics.
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