$\left(2a^{-1}+3b^{-1}\right)^{-1}$
$=\left(2\cdot\frac{1}{a}+3\cdot\frac{1}{b}^{}\right)^{-1}$
$=\left(\frac{2}{a}+\frac{3}{b}^{}\right)^{-1}$
$=\left(\frac{2b+3a}{ab}^{}\right)^{-1}$
$=\frac{1}{\frac{2b+3a}{ab}}$
$=1\times\frac{ab}{2b+3a}$
$=\frac{ab}{3a+2b}$ [Answer]
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