Question

已知A+B=1，A的平方+B的平方=2，求A的七次方+B的七次方的值

Answer 1

a+b =1 两边平方得
a^2+b^2+2ab = 1
解得
ab =-1/2
则
(ab)^2 = 1/4
(ab)^4 = 1/16
a^4+b^4 = 4-2(ab)^2 = 4-1/2 = 7/2
而
a^7+b^7 = (a+b)(a^7+b^7) = a^8+b^8 + ab^7+ba^7

= (a^8+b^8)+ab(a^6+b^6)

= (a^4+b^4)^2-2(ab)^4+ab[(a^2+b^2)(a^4+b^4-(ab)^2)]

= (7/2)^2 - 2*(1/16)+(-1/2)*[2*(7/2 - 1/4)]

= 71/8

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解法2：
a+b =1 两边平方得
a^2+b^2+2ab = 1
解得
ab =-1/2
(ab)^2 = 1/4
(ab)^3 = -1/8
而
a^3+b^3 = (a+b)(a^2-ab+b^2) = a^2-ab+b^2 = 2+1/2 = 5/2
a^4+b^4 = (a^2+b^2)^2 - 2(ab)^2 = 4 -1/2 = 7/2

所以
a^7+b^7 = (a^3+b^3)(a^4+b^4)-a^3b^4-b^3a^4 = (a^3+b^3)(a^4+b^4)-(ab)^3(a+b) = (5/2)*(7/2)-(-1/8)*1 = 71/8

希望可以采纳

Answer 2

A+B=1
A^2+B^2+2AB=1
A^2+B^2=2
2AB=-1
A=-1,B=1/2
A^7+B^7=(-1)^7+(1/2)^7
=-1+1/128
=-127/128