化简 $\dfrac{x^2-9y^2}{3x^2+7xy-6y^2}$。

1. Factorise the numerator as a difference of two squares.

$x^2-9y^2=x^2-(3y)^2=(x+3y)(x-3y)$

2. Factorise the quadratic denominator.

$3x^2+7xy-6y^2$
Find two terms whose product is $3x^2(-6y^2)=-18x^2y^2$ and whose sum is $7xy$: $9xy$ and $-2xy$.
$3x^2+7xy-6y^2=3x^2+9xy-2xy-6y^2$
$=3x(x+3y)-2y(x+3y)$
$=(x+3y)(3x-2y)$

3. Cancel the common factor and write the simplified expression.

$\frac{x^2-9y^2}{3x^2+7xy-6y^2}=\frac{(x+3y)(x-3y)}{(x+3y)(3x-2y)}$
Cancel the common factor $(x+3y)$:
$=\frac{x-3y}{3x-2y}$