Đặt \(\dfrac{x}{18}=\dfrac{y}{19}=k\Rightarrow\left\{{}\begin{matrix}x=18k\\y=19k\end{matrix}\right.\left(1\right)\)
Thay (1) vào biểu thức đã cho, ta có:
\(\dfrac{2x^2-3y^2}{2x+3y}=\dfrac{2\left(18k\right)^2-3\left(19k\right)^2}{2\cdot18k+3\cdot19k}\\ =\dfrac{648k-1083k}{36k+57k}\\ =\dfrac{-435k}{93k}\\ =\dfrac{-145}{31}\)