[tex]\dfrac{\frac{x+2}{x^2+2x-3}}{\frac{x+2}{x^2-x}}[/tex]
If [tex]x\neq-2[/tex], then we can immediately cancel the factors of [tex]x+2[/tex]:
[tex]\dfrac{\frac1{x^2+2x-3}}{\frac1{x^2-x}}=\dfrac{x^2-x}{x^2+2x-3}[/tex]
Factorize the numerator and denominator:
[tex]x^2-x=x(x-1)[/tex]
[tex]x^2+2x-3=(x+3)(x-1)[/tex]
Next, if [tex]x\neq1[/tex], then
[tex]\dfrac{x^2-x}{x^2+2x-3}=\dfrac{x(x-1)}{(x+3)(x-1)}=\dfrac x{x+3}[/tex]
