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解答:
由已知:x^2+√3y=√5,y^2+√3x=√5,所以:x^2+√3y=y^2+√3x
即:x^2-y^2+√3y-√3x=0,所以:(x+y)(x-y)-√3(x-y)=0
因为:x≠y,所以:x+y=√3,
则:y=√3-x,x=√3-y,
分别代入x^2+√3y=√5,y^2+√3x=√5中,得到:x^2-√3x+3-√5=0,y^2-√3y+3-√5=0
即:x,y分别为方程A^2-√3A+3-√5=0的两个根,
所以:x+y=√3,xy=3-√5,
所以:y/x+x/y=(x^2+y^2)/(xy)=[(x+y)^2-2xy]/(xy)=(x+y)^2/(xy)-2=3/(3-√5)-2
=3/4*(3+√5)-2=(3√5+1)/4。
由已知:x^2+√3y=√5,y^2+√3x=√5,所以:x^2+√3y=y^2+√3x
即:x^2-y^2+√3y-√3x=0,所以:(x+y)(x-y)-√3(x-y)=0
因为:x≠y,所以:x+y=√3,
则:y=√3-x,x=√3-y,
分别代入x^2+√3y=√5,y^2+√3x=√5中,得到:x^2-√3x+3-√5=0,y^2-√3y+3-√5=0
即:x,y分别为方程A^2-√3A+3-√5=0的两个根,
所以:x+y=√3,xy=3-√5,
所以:y/x+x/y=(x^2+y^2)/(xy)=[(x+y)^2-2xy]/(xy)=(x+y)^2/(xy)-2=3/(3-√5)-2
=3/4*(3+√5)-2=(3√5+1)/4。
2011-03-05 · 知道合伙人教育行家
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x^2+根号3 y=根号5......(1)
y^2+根号3 x=根号5......(2)
(1)-(2)得:
(x^2-y^2)-根号3(x-y)=0
(x+y)(x-y)-根号3(x-y)=0
(x-y)(x+y-根号3)=0
x≠y,即x-y≠0
∴x+y-根号3=0
∴x+y=根号3
(1)+(2)得:
x^2+y^2+根号3(x+y)=2根号5
(x+y)^2-2xy+根号3(x+y)=2根号5
(根号3)^2-2xy+根号3*根号3=2根号5
2xy=6-2根号5
xy=3-根号5
y/x+x/y=(x^2+y^2)/(xy)
={(x+y)^2-2xy}/(xy)
=(x+y)^2/(xy) - 2
=(根号3)^2/(3-根号5) - 2
=3(3+根号5)/(9-5) - 2
=(9+3根号5)/4-2
=(1+3根号5)/4
y^2+根号3 x=根号5......(2)
(1)-(2)得:
(x^2-y^2)-根号3(x-y)=0
(x+y)(x-y)-根号3(x-y)=0
(x-y)(x+y-根号3)=0
x≠y,即x-y≠0
∴x+y-根号3=0
∴x+y=根号3
(1)+(2)得:
x^2+y^2+根号3(x+y)=2根号5
(x+y)^2-2xy+根号3(x+y)=2根号5
(根号3)^2-2xy+根号3*根号3=2根号5
2xy=6-2根号5
xy=3-根号5
y/x+x/y=(x^2+y^2)/(xy)
={(x+y)^2-2xy}/(xy)
=(x+y)^2/(xy) - 2
=(根号3)^2/(3-根号5) - 2
=3(3+根号5)/(9-5) - 2
=(9+3根号5)/4-2
=(1+3根号5)/4
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