4) 3n²-3m²/n²+np · 6m-6n/n+p ;

$$\frac{3n^2-3m^2}{n^2+np} \cdot \frac{6m-6n}{n+p} = \frac{3(n^2-m^2)}{n(n+p)} \cdot \frac{6(m-n)}{n+p} = \frac{3(n-m)(n+m)6(m-n)}{n(n+p)^2} = \frac{-18(n-m)^2(n+m)}{n(n+p)^2}$$

Ответ: $$\frac{-18(n-m)^2(n+m)}{n(n+p)^2}$$