23. If $x$ is positive, find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$. If $x$ is positive, find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$.
23. $\sqrt{10}$
23. Let $A, B, P$ be the points $(0,-1),(1,2)$ and $\left(x, \frac{1}{x}\right)$ respectively. Then
$$
\begin{aligned}
\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x} & =\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}}+\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\
& =\sqrt{x^{2}+\left(\frac{1}{x}+1\right)^{2}}+\sqrt{(x-1)^{2}+\left(\frac{1}{x}-2\right)^{2}} \\
& =P A+P B \\
& \geq A B \\
& =\sqrt{(0-1)^{2}+(-1-2)^{2}} \\
& =\sqrt{10}
\end{aligned}
$$

Equality is possible if $A, P, B$ are collinear, i.e. $P$ is the intersection of the curve $x y=1$ with the straight line $A B$ in the first quadrant. One can find by simple computation that $x=\frac{1+\sqrt{13}}{6}$ in this case. Hence the answer is $\sqrt{10}$.