Question

在线求解数学题

已知：a+b+c=0 求：a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)的值

Answer 1

解：
因为a+b+c=0，
所以a+b=－c，a+c=－b，b+c=－a

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+c)/b+(a+b)/c+(b+c)/a

=(－b/b)+(－c/c)+(－a/a)

=－3

Answer 2

由a+b+c=0 ，得 a+b=-c，a+c=-b，b+c=-a （后面要用）

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

Answer 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/b+a/c+b/a+b/c+c/a+c/b+(a/a+b/b+c/c)-3
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c-3
=-3

Answer 4

解: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
=1+(b+c)/a+1+(a+c)/b+1+(a+b)/c-3 (注:1=a/a 1=b/b 1=c/c)
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c-3
把a+b+c=0代入上式得:
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3

Answer 5

由题意，a,b,c都不为0。所以化解成(a+c)/b+(a+b)/c+(b+c)/a=-1-1-1=-3

Answer 6

对a+b+c=0分析
两边同时乘以a有：aa+ab+ac=0
再两边乘以c有:aac+abc+acc=0
移项： aac+acc=-abc
同理可证：aab+bba=-abc
bbc+ccb=-abc
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
=(aac+aab+bbc+bba+ccb+cca)/abc
=(-3abc)/abc=-3