Rational

$\frac{x+2}{4x+8}-\frac{3-x}{x^2+2x}$

= $\frac{(x+2)(x^2+2x)}{(4x+8)(x^2+2x)}-\frac{(3-x)(4x+8)}{(x^2+2x)(4x+8)}$

= $\frac{x^3+4x^2+4x}{(4x+8)(x^2+2x)}-\frac{-4x^2+4x+24}{(x^2+2x)(4x+8)}$

= $\frac{x^3+8x^2-24}{(x^2+2x)(4x+8)}$

= $\frac{(x+2)(x^2+6x-12)}{x(x+2)(4x+8)}$

= $\frac{x^2+6x-12}{4x^2+8x}$

Restrictions: X cannot equal 0, -2, as these values would make the denominator 0.