Question

已知x+y=1,x^3+y^3=1/3,求 x^5+y^5

Answer 1

答：
x^3+y^3
=(x+y)(x^2-xy+y^2)
=x^2-xy+y^2=1/3
(x+y)^2=x^2+2xy+y^2=1
所以相减得xy=2/9
x^5+y^5
=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
=x^4-x^3y+x^2y^2-xy^3+y^4
=(x^2-xy+y^2)^2+x^3y+xy^3-2x^2y^2
=1/9+xy(x^2-xy+y^2)-x^2y^2
=1/9+2/27-4/81
=11/81

Answer 2

立方得：(x y)^3=1
x^3 y^3 3xy(x y)=1
3 3xy=1
所以xy=-2/3,
故 x^2 y^2=(x y)^2-2xy=1 4/3=7/3
又(x^3 y^3)(x^2 y^2)=7=x^5 y^5 x2y^2(x y)=x^5 y^5 4/9
所以x^5 y^5=59/9『望采纳』

追问

呵 你这是抄来的吧 我是三分之一 不是三！！！