作业帮分式加减 (22 17:21:41)abc不等于0,a+b+c=0,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)等于多少?
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分式加减 (22 17:21:41)
abc不等于0,a+b+c=0,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)等于多少?
-3
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
=(a+c)/b+(a+b)/c+(b+c)/a
因为 abc不等于0,a+b+c=0
所以 a+c=-b
a+b=-c
b+c=-a
所以(a+c)/b+(a+b)/c+(b+c)/a=-1+（-1）+(-1)=-3
即a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3
=a/b+a/c+b/a+b/c+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)/a
=-3
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^3/(abc)-b^3/(abc)-c^3/(abc)
=((b+c)^3-b^3-c^3)/(-bc(b+c))
=(3b^2c+3bc^2)/(-bc(b+c))
=-3