谁能回答我的数学问题

a+b=1
a^2+b^2=2
(a+b)^2-2ab=2
ab=-1/2
a^7+b^7
=(a^3+b^3)*(a^4+b^4)-(a^3b^4+a^4b^3)
=(a+b)(a^2-ab+b^2)[(a^2+b^2)^2-2a^2b^2]-a^3b^3(a+b)
=1*(2+1/2)(2^2-2*1/4)+1/8*1
=35/4+1/8
=71/8

我能

解：由题意可知
a+b=1
a^2+b^2=2
即ab=[(a+b)^2-(a^2+b^2)]/2=-1/2
所以a^3+b^3=(a+b)(a^2-ab+b^2)=5/2
a^4+b^4=(a^2+b^2)^2-2(ab)^2=7/2
a^7+b^7=(a^3+b^3)(a^4+b^4)-(ab)^3(a+b)=5*7/4+1/8=71/8

已知:a+b=1,a^2+b^2=2
则(a+b)^2=1,即a^2+b^2+2ab=1,求得ab=-1/2
1.(a^2+b^2)(a+b)=2x1=2
a^3+b^3+ab(a+b)=2,得a^3+b^3=5/2
以此类推
2.(a^3+b^3)(a+b)=5/2x1=5/2
a^4+b^4+ab(a^2+b^2)=5/2,得a^4+b^4=7/2

3.(a^4+b^4)(a+b)=7/2x1=7/2
a^5+b^5+ab(a^3+b^3)=7/2,得a^5+b^5=19/4

4.(a^5+b^5)(a+b)=19/4x1=19/4
a^6+b^6+ab(a^4+b^4)=19/4,得a^6+b^6=13/2

5.(a^6+b^6)(a+b)=13/2x1=13/2
a^7+b^7+ab(a^5+b^5)=13/2,得a^7+b^7=71/8