\(已知2^a=6^b=144,求\sqrt{\frac{1}{a}+\frac{1}{b}}\)
\[
\begin{align*}
& 2^a = 144 \\
& (2^a)^{\frac{1}{a}} = 144^{\frac{1}{a}} 令为1式。\\
& (6^b)^{\frac{1}{b}} = 144^{\frac{1}{b}} 令为2式。\\
& 1式 \cdot 2式 \\
& 2 \cdot 6 = 144^{\frac{1}{a}+\frac{1}{b}} \\
& 解得: \frac{1}{a}+\frac{1}{b} = \frac{1}{2} \\
& 即求：\sqrt{\frac{1}{a}+\frac{1}{b}} = \sqrt{\frac{1}{2}}\\
& =\frac{\sqrt{2}}{2}
\end{align*}
\]