Đặt $\dfrac ab = \dfrac cd = k$
$\to \begin{cases}a = kb\\c = kd\end{cases}$
Ta được:
$+) \quad \dfrac{3a + 2c}{3b + 2d}$
$=\dfrac{3.kb + 2.kd}{3b + 2d}$
$= \dfrac{k(3b + 2d)}{3b + 2d} = k$
$+) \quad \dfrac{5a + 3c}{5b + 3d}$
$=\dfrac{5.kb + 3.kd}{5b+3d}$
$=\dfrac{k(5b + 3d)}{5b + 3d} = k$
Do đó:
$\dfrac{3a + 2c}{3b + 2d} = \quad \dfrac{5a + 3c}{5b + 3d}$