Answer:
Remember, [tex]\frac{a}{b}\div \frac{c}{d}=\frac{a}{b}*\frac{d}{c}=\frac{a*c}{b*d}[/tex]
Then,
[tex]\frac{x+1}{x+2}\div\frac{3x}{x-4}=\frac{x+1}{x+2}*\frac{x-4}{3x}=\frac{(x+1)(x-4)}{3x(x+2)}=\frac{x^2-3x-4}{3x^2+6x}[/tex],
Now, using long division of polynomies we have that
[tex]\frac{x^2-3x-4}{3x^2+6x}=\frac{1}{3}-\frac{5x+4}{3x^2+6x}[/tex]
