Đặt $\dfrac{a}{b} = \dfrac{c}{d} = k ( k \ne 0 )$
$\Rightarrow \begin{cases} a = kb \\ c = kd \end{cases}$
Khi đó :
$\dfrac{7a^2 - 4ab +2b^2}{b^2 +3ab} = \dfrac{7k^2b^2 - 4kb^2 + 2b^2}{b^2 + 3kb^2} = \dfrac{b^2(7k^2 - 4k + 2)}{b^2(1 +3k)} = \dfrac{7k^2 - 4k +2}{1+3k}$
$\dfrac{7c^2 - 4cd +2d^2}{d^2 +3cd} = \dfrac{7k^2d^2 - 4kd^2 + 2d^2}{d^2 + 3kd^2} = \dfrac{d^2(7k^2 - 4k + 2)}{d^2(1 +3k)} = \dfrac{7k^2 - 4k +2}{1+3k}$
$\Rightarrow \dfrac{7a^2 - 4ab +2b^2}{b^2 +3ab} = \dfrac{7c^2 - 4cd +2d^2}{d^2 +3cd}$