设x+y=1，x^2+y^2=2，求x^7+y^7的值.

来源：百度知道编辑：UC知道时间：2024/06/22 20:57:46

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

x+y=1
两边平方
x^2+y^2+2xy=1
xy=[1-(x^2+y^2)]/2=-1/2
所以x^4+y^4=(x^2+y^2)^2-2(xy)^2=4-2*(-1/2)^2=7/2

所以x^7+y^7
=(x^3+y^3)(x^4+y^4)-x^3y^4-x^4y^3
=(x^3+y^3)(x^4+y^4)-(xy)^3(x+y)
=1*7/2-(-1/2)^3*1
=29/8

得仔细算

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