1/n(n-1)+1/n(n+1)=(n+1+n-1)/[n(n²-1)]=(2n/[n(n²-1)]=2/(n²-1)
2/(n²-1)8* (n²-1)/(n+3)=2/(n+3)
1)sin^2x+2sinx*cosx-3cos^2x=0 |÷cos^2x
tg^2x+2tgx-3=0
tgx=t
t^2+2t-3=0
x1=-3(не верно) ,x2=1
tgx=1
x=πn,n€Z
2)cos5x=-1/2
5x=+-2π/3+2πn. ,|÷5
x=+-2π/15+2πn/5, n€Z