已知x=1⼀2(根号7+根号5),y=1⼀2(根号7-根号5),求x²-xy+y²的值

2024-12-26 02:13:34
推荐回答(3个)
回答1:

x²-xy+y²
=x²-2xy+y²+xy
=(x-y)²+xy
=[1/2(根号7+根号5)-1/2(根号7-根号5)]²+1/2(根号7+根号5)*1/2(根号7-根号5),
=[(√7+√5-√7+√5)/2]²+(7-5)/4
=(√5)²+1/2
=5+1/2
=11/2
=5.5

回答2:

x²-xy+y²
=(x+y)²-3xy
=[1/2(根号7+根号5)+1/2(根号7-根号5)]²-3×1/2(根号7+根号5)×1/2(根号7-根号5)
=7-3/2
=5又1/2

回答3:

(x-y)²+xy=根号5的平方+1/4(根号7+根号5)(根号7-根号5)
=5+1/4(7-5)
=5.5