已知x+y=5,xy=2,求下列各式的值:(1)x2+y2 ;(2)(x-y)2. x8+y8的值是多少!

2024-12-29 08:55:41
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回答1:

解答:
x^2+y^2=(x+y)^2-2xy=25-4=21

(x-y)^2=(x+y)^2-4xy=25-8=17

x^8+y^8=(x^4+y^4)^2-2x^4y^4=[(x^2+y^2)^2-2x^2y^2]^2-2(xy)^4
={[(x+y)^2-2xy]^2-2(xy)^2]^2-2(xy)^4
=(21^2-8)^2-2*16=187457