срочно!!!!!
надо решить пример

1)1/3(n-1)-2/(n-3)(n+3)-1/n(n+3)=[n(n+3)-2*3n-3(n-3)]/3n(n-3)(n+3)=
=(n²+3n-6n-3n+9)/3n(n-3)(n+3)=(n²-6n+9)/3n(n-3)(n+3)=
=(n-3)²/3n(n-3)(n+3)=(n-3)/3n(n+3)
2)9(n+3)²/(n-3) * (n-3)/3n(n+3)=3(n+3)/n
3)(9n+27)/n²(3-n)+3(n+3)/n=(9n+27+3n(3-n)(n+3)/n²(3-n)=
=(9n+27+27n-3n³)/n²(3-n)=-3(n²-12n-9)/[n²(3-n)]=3(n²-12n-9)/[n²(n-3)]