(3y²+5)(y-6)=3y³+5y-18y²-30=3y³-18y²+5y-30
(7c²-1)(c-3)=7c³-c-21c²+3=7c³-21c²-c+3
Ctgx-tgx=1/tgx-tgx=(1-tg²x)/tgx=2*(1-tg²x)/2tgx=2/tg2x
2*(1/(tgx+1) +1/(tgx-1))=2*(tgx-1+tgx+1)/(tg²x-1))=2*2tgx/(tg²x-1)=-2tg2x
2/tg2x -2tg2x=4
2-2tg²2x-4tg2x=0 tg2x≠0
tg²2x+2tg2x-1=0
tg2x=a
a²+2a-1=0
D=4+4=8
a1=(-2-2√2)/2=-1-√2⇒tg2x=-1-√2⇒2x=-arctg(1+√2)+πn⇒x=-1/2arctg(1+√2)+πn/2
a2=(-2+2√2)/2=√2-1⇒tg2x=√2-1⇒2x=arctg(√2-1)+πn⇒x=1/2arctg(√2-1)+πn/2
По теореме Виета сумма корней квадр. уравнения = -18/2=-9
(-2х³у)²=(-2х³)²*(у)²=4х^6y²