初二数学！！若a+b+c=0,且abc≠0,求a(1⼀b+1⼀c)+b(1⼀a+1⼀c)+c(1⼀a+1⼀b)的值

a{1/b+1/c}+b{1/c+1/a}+c{1/a+1/b}
=a/b+a/c+b/c+b/a+c/a+c/b
=[a(b+c)+b(a+c)+c(a+b)]/abc
=(2ab+2ac+2bc)/abc..........①

∵a+b+c=0;
∴(a+b+c)(a+b+c)=a*a+b*b+c*c+2ab+2ac+2bc=0;
∴2ab+2ac+2bc=0
∴①=0/abc
=0

2.(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到：
XYZ+1/(XYZ)=2,解得XYZ=1

1.
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a/b+a/c+b/a+b/c+c/a+c/b
=(b+c)/a+(a+c)/b+(a+b)/c
=-a/a-b/b-c/c
=-1-1-1
=-3
2.
∵x+1/y=4 ∴x=(4y-1)/y
∵y+1/z=1 ∴z=1/(1-y)
z+1/x=7/3
∴1/(1-y)+y/(4y-1)=7/3
y=2/5
∴x=3/2,z=5/3
∴xyz=3/2*2/5*5/3
=1

∵a+b+c=0;
∴(a+b+c)(a+b+c)=a*a+b*b+c*c+2ab+2ac+2bc=0;
∴2ab+2ac+2bc=0
∴①=0/abc
=0
是0!!!!!!!!!!!!!!!!11

你是错的真确的

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=a(1/a+1/b+1/c) + b(1/b+1/c+1/a) + c(1/c+1/a+1/b) -3
=(a+b+c)*(1/a+1/b+1/c)-3
=-3

1）
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a(b+c)/bc+b(a+c)/ac+c(a+b)/ab
=-a^2/bc-b^2/ac-c^2/ab
=-(a^3+b^3+c^3)/abc
=-[(a^3+b^3+c^3-3abc)/abc+3abc/abc]
=-[(a+b+c)(a^2+b^2+c^2-ab-bc-ca)/abc+3]
=-[0+3]
=-3