已知a+b=1,a²＋b²=2,那么a*a*a*a*a*a*a+b*b*b*b*b*b*b=?

Question

如题所述

Answer 1

∵a+b=1,a²＋b²=2,
∴（a+b)²=1
∴a²+2ab+b²=1
∴2ab+2=1
∴ab=-1/2

(a^2+b^2)^2=4
a^4+b^4+2(ab)^2=4
a^4+b^4=3.5
(a^4+b^4)^2=12.25
a^8+b^8+2(ab)^4=12.25
a^8+b^8=12.125

a^6+b^6
=(a^2+b^2)[a^4+b^4-(ab)^2]
=2*(3.5-0.25)
=6.5

a^7+b^7
=(a^7+b^7)*(a+b)
=a^8+b^8+ab^7+ba^7
=(a^8+b^8)+ab(a^6+b^6)
=12.125+(-0.5)*6.5
=8.875

Answer 2

a+b=1,a²＋b²=2

a²+2ab+b²=1
ab=-1/2
a³+b³=(a+b)³-3ab(a+b)=1+3/2=5/2
a^4+b^4+2a²b²=4
a^4+b^4=4-1/2=7/2

a^7+b^7=(a^4+b^4)(a^3+b^3)-a^4b^3-a^3b^4
=(a^4+b^4)(a^3+b^3)-a³b³(a+b)
=7/2*5/2-(-1/2)³
=35/4+1/8
=71/8本回答被提问者采纳