x=1/(根号3-2),y=1/(根号3+2),求代数式(x2+xy+y2)/(x+y)的值
问题描述：
x=1/(根号3-2),y=1/(根号3+2),求代数式(x2+xy+y2)/(x+y)的值
答
x=1/(√3-2)=(√3+2)/[(√3-2)(√3+2)]=-(√3+2)
y=1/(√3+2)=(√3-2)/[(√3+2)(√3-2)]=-(√3-2)
x+y=-2√3
xy=-1
(x^2+xy+y^2)/(x+y)
=[(x+y)^2-xy]/(x+y)
=[(-2√3)^2+1]/(-2√3)
=13/(-2√3)
=-13√3/6
答
x=1/(√3-2)=(√3+2)/(√3-2)(√3+2)=-(√3+2)
y=1/(√3+2)=(√3-2)/(√3-2)(√3+2)=2-√3
x+y=-(√3+2)+2-√3=-2√3
xy=-(√3+2)(2-√3)=-1
(x²+xy+y²)/(x+y)
=［(x+y)²-xy］/(x+y)
=(12+1)/(-2√3)
=-13√3/6
答
x=1/(√3-2)=-2-√3
y=1/(√3+2)=2-√3
x+y=-2√3
xy=-1
(x2+xy+y2)/(x+y)=[(x+y)²-xy]/(x+y)=11/(-2√3)=-11√3/6