Question

已知:x=根号3+1,y=根号3-1,求x^2-2xy+y^2分之x^2-y^2

如题，求速度！

Answer 1

解：x＝√3＋1, y＝√3－1
x－y＝√3＋1－√3＋1＝2
x＋y＝√3＋1＋√3－1＝2√3
∴ ﹙x²－y²﹚/﹙x²－2xy＋y²﹚
＝﹙x－y﹚﹙x＋y﹚/﹙x－y﹚²
＝﹙x＋y﹚/﹙x－y﹚
＝2√3/2
＝√3.

Answer 2

(x²-y²)/(x²-2xy+y²)
=(x+y)(x-y)/(x-y)²
=(x+y)/(x-y)
当x=根号3+1,y=根号3-1时
原式=（√3+1+√3-1)/(√3+1-√3+1）=2√3/2=√3
希望采纳！

Answer 3

x=√3+1,y=√3-1
(x^2-y^2)/(x^2-2xy+y^2)
=(x-y)(x+y)/(x-y)^2
=(x+y)/(x-y)
=(√3+1+√3-1)/[√3+1-(√3-1)]
=2√3/[√3+1-√3+1)
=2√3/2
=√3

Answer 4

（x^2-2xy+y^2）/（x^2-y^2）=（x-y)^2/(x-y)(x+y)=(x-y)/（x+y)=(根号3+1-根号3+1)/(根号3+1+根号3-1)=2/2根号3=1/根号3
=根号3/3