Question

abc≠0,a+b+c=0求a²/bc+b²/ac+c²/ab=?

Answer 1

原式=a^3/abc+b^3/abc+c^3/abc
=(a^3+b^3+c^3)/abc
=(a^3+b^3+c^3-3abc)/abc+3
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)/abc+3
=3

Answer 2

3
a²/bc+b²/ac+c²/ab通分得(a^3+b^3+c^3)/abc
因为2(a^3+b^3+c^3)=(a^3+b^3)+(a^3+c^3)+(b^3+c^3)
=(a+b)(a^2+b^2-ab)+(a+c)(a^2+c^2-ac)+(b+c)(b^2+c^2-bc)
=-c(a^2+b^2-ab)-b(a^2+c^2-ac)-a(b^2c^2-bc)
=-a^2 c-b^2 c+abc-a^2 b-c^2 b+abc-b^2 a-c^2 a+abc
=-a^2(b+c)-b^2(a+c)-c^2(a+B)+3abc
=-a^2(-a)-b^2(-b)-c^2(-c)+3abc
=a^3+b^3+c^3+3abc
即2(a^3+b^3+c^3)=a^3+b^3+c^3+3abc
所以有a^3+b^3+c^3=3abc
原题变为:
3abc/abc=3

Answer 3

a²/bc+b²/ac+c²/ab=(a^3+b^3+c^3)/(abc)
a^3+b^3=(a+b)(a^2-ab+b^2)
a^2-ab+b^2=(a+b)^2-3ab=c^2-3ab
a^3+b^3=(a+b)(a^2-ab+b^2)=-c(c^2-3ab)=-c^3+3abc
a^3+b^3+c^3=3abc
a²/bc+b²/ac+c²/ab=(a^3+b^3+c^3)/(abc)=3

Answer 4

a²/bc+b²/ac+c²/ab=(a^3+b^3+c^3)/(abc)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)/(abc)
=0

Answer 5

同乘abc 式为a3+b3+c3=?
a=-(b+c)代入为-3bc(b+c)=3abc
需要具体数字吗