X²+5x-16=0
по теореме Виета:
x²+px+q=0;⇒p=-(x₁+x₂);q=x₁·x₂;⇒
p=5;q=-16;
1)x₁²x₂+x₂²x₁x=x₁·x₂(x₁+x₂)=q·(-p)=(-16)·(-5)=80;
2)x₂/x₁+x₁/x₂=(x₂x₂+x₁x₁)/(x₁x₂)=[(x₂²+x₁²+2x₁x₂)-2x₁x₂]/x₁x₂=
=[(x₁+x₂)²-2x₁x₂]/x₁x₂=[(-p)²-2q]/q=(5²-2·(-16))/16=(25+32)/(-16)=
=-57/16=-3.5625;