Question

代数求值1、已知abc≠0,且a+b+c=0,则代数式a²/（bc）+b²/（ac）+c²

Answer 1

1.abc≠0说明a,b,c没有一个是0,但a+b+c=0,所以a=b=c=0

所以a²/（bc）+b²/（ac）+c²/（ab)=0

2.∵a^2/(a^4+a^2+1)=1/24

∴(a^4+a^2+1)/a^2=24

a^2+1+(1/a^2)=24

a^2+2(a)×(1/a)+(1/a^2)=25

[a+(1/a)]^2=25

a+(1/a)=5 应该有a＞0的条件

∵b^3/(b^6+b^2+1)=1/19

∴(b^6+b^3+1)/b^3=19

b＾3+(1/b＾3)+1=19

[b+(1/b)][b＾-1+(1/b＾)]=18

[b+(1/b)]{[b+(1/b)]＾-3]=18

b+(1/b)=u

u(u＾-3)=18

u＾3-3u-18=0

u＾3-3u＾+3u＾-3u-18=0

(u＾)(u-3)+3(u＾-u-6)=0

(u＾)(u-3)+3(u+2)(u-3)=0

(u-3)(u＾+3u+6)=0

∵(u＾+3u+6)=0时 △=9-24＜0无实数根

∴u=3=0 u=3

∴ab/(a^2+a+1)(b^2+b+1)

=1/[a+(1/a)+1][b+(1/b)+1]

=1/(6×4)=1/24