Question

a+b+c=0,求a²/(2a²+bc)+b²/(2b²+ac)+c²/(2c²+ab)的值

详细过程

Answer 1

解:因为a+b+c=0
有a=-(b+c) b=-(c+a) c=-(a+b)
则原式=a方/(2a方+bc)+b方/(2b方+ac)+c方/(2c方+ab)
=a^2/(2a^2+bc)+b^2/(2b^2+ac)+c^2/(2c^2+ab)
=a^2/[a^2-a(b+c)+bc]+b^2/[b^2-b(a+c)+ac]+c^2/[c^2-c(a+b)+ab]
=a^2/[a(a-c)-b(a-c)]+b^2/[b(b-c)-a(b-c)]+c^2/[c(c-b)-a(c-b)]
=a^2/[(a-b)(a-c)]+b^2/(b-a)(b-c)+c^2/[(c-a)(c-b)]
=1/(a-b)*[a^2/(a-c)-b^2/(b-c)]+c^2/[(a-c)(b-c)
=1/(a-b)*(a^2b^2-ab^2-a^2c+b^2c)/[(a-c)(b-c)+c^2/[(a-c)(b-c)
=(ab-ca-cb)/[(a-c)(b-c)+c^2/[(a-c)(b-c)
=(ab-ca-cb+c^2)/[(a-c)(b-c)
=(a-c)(b-c)/[(a-c)(b-c)]
=1

Answer 2

收藏！

Answer 3

869

追问

详细过程

Answer 4

-0