5・146 Find integer coefficient polynomials $P(x)$ and $Q(x)$, such that
$$\frac{P(\sqrt{2}+\sqrt{3}+\sqrt{5})}{Q(\sqrt{2}+\sqrt{3}+\sqrt{5})}=\sqrt{3}+\sqrt{2}$$
[Solution] Let $S=\sqrt{2}+\sqrt{3}+\sqrt{5}$. We first find polynomials $A(x)$ and $Q(x)$ such that $\frac{A(S)}{Q(S)}=\sqrt{5}$. For this, we consider
$$\begin{array}{l}
x-\sqrt{2}-\sqrt{3}-\sqrt{5}, x+\sqrt{2}+\sqrt{3}-\sqrt{5} \\
x+\sqrt{2}-\sqrt{3}-\sqrt{5}, x-\sqrt{2}+\sqrt{3}-\sqrt{5}
\end{array}$$

Their product equals $x^{4}-4 \sqrt{5} x^{3}+20 x^{2}-24$. Therefore, if we let
$$A(x)=x^{4}+20 x^{2}-24, Q(x)=4 x^{3}$$

then we have $A(S)-\sqrt{5} Q(S)=0$,
i.e., $\square$
$$\begin{array}{l}
\frac{A(S)}{Q(S)}=\sqrt{5} \\
\frac{S^{4}+20 S^{2}-24}{4 S^{3}}=\sqrt{5}
\end{array}$$

Thus, we get $\sqrt{2}+\sqrt{3}=S-\sqrt{5}=\frac{3 S^{4}-20 S^{2}+24}{4 S^{3}}$. This means we can take
$$\begin{array}{l}
P(x)=3 x^{4}-20 x^{2}+24, Q(x)=4 x^{3} \text { such that } \\
\frac{P(S)}{Q(S)}=\sqrt{3}+\sqrt{2} .
\end{array}$$