Example 10. Simplify: $\sqrt{(-45)^{2}}+\frac{1}{\sqrt{12-\sqrt{140}}}$
$$
-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}} .
$$
$$
\begin{array}{l}
-\frac{1}{\sqrt{8+\frac{v^{\prime} 64-60}{2}}}-\sqrt{\frac{8-\sqrt{64}-60}{2}} \\
2 \\
\sqrt{\frac{10+\sqrt{100-84}}{2}}+\sqrt{\frac{10-\sqrt{100-84}}{2}} \\
=3 \sqrt{5} \text {. } \\
\end{array}
$$

This problem uses the following theorem:
If $a>0, b>0$, and $a^{2}>b$, then
$$
\sqrt{a \pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^{2}}-b}{2}} .
$$

We provide the proof as follows.
$$
\begin{array}{l}
\text { Let } \sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=x, (x>0) \\
\text { Then } 2 a+2 \sqrt{a^{2}-b}=x^{2} \text {, } \\
\therefore x=\sqrt{2 a+2 \sqrt{a^{2}-b}}=2 \sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}, \\
\text { i.e., } \sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=2 \sqrt{\frac{a+\sqrt{a^{2}-b}}{2}} \text {. } \\
\end{array}
$$

Similarly, we get $\sqrt{a+\sqrt{b}}-\frac{1}{a}-\sqrt{\sqrt{b}}$
$$
=2 \sqrt{\frac{a}{d}=\frac{b}{2}} \text {. }
$$

Solving (1) and (2) yields
$$
\begin{array}{c}
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{a^{2}-b}}{2}} \\
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{a^{2}-b}}{2}} . \\
\text { i.e., } \sqrt{a \pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^{2}-b}}{2}} .
\end{array}
$$

Special case:
If $a>b>0$, then $\sqrt{(a+b) \pm 2 \sqrt{a b}}$
$$
=\sqrt{(\sqrt{a} \pm \sqrt{b})^{2}}=\sqrt{a} \pm \sqrt{b} .
$$