Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.


Solve the conjugate expression
$$
y=(15-\sqrt{220})^{19}+(15-\sqrt{220})^{82} \text {. }
$$

Then $x+y$
$$
\begin{aligned}
= & (15+\sqrt{220})^{19}+(15-\sqrt{220})^{19}+ \\
& (15+\sqrt{220})^{82}+(15-\sqrt{220})^{82} .
\end{aligned}
$$

By the binomial theorem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
(15+\sqrt{220})^{n}+(15-\sqrt{220})^{n} \\
=2\left(\mathrm{C}_{n}^{0} 15^{n}+\mathrm{C}_{n}^{2} 15^{n-2} \times 220+\cdots\right) \in \mathbf{Z}_{+},
\end{array}
$$

and the last digit is zero.
Therefore, $x+y$ is a positive integer with the last digit being zero. Also, $0<15-\sqrt{220}=\frac{5}{15+\sqrt{220}}<\frac{5}{25}=\frac{1}{5}$, and $(15-\sqrt{220})^{82}<(15-\sqrt{220})^{19}$, so $0<y<2(15-\sqrt{220})^{19}<2 \times 0.2^{19}<0.4$. Therefore, the last digit of $x$ is 9.