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a+b+c=0则a(1除以b+1除以c)+b(1除以a+1除以c?

2012-04-24 19:43:59圆***
a+b+c=0,则a(1除以b+1除以c)+b(1除以a+1除以c)+c(1除以a+1除以b)的值是谢谢了 a+b+c=0则a(1除以b+1除以c)+b(1除以a+1除以c)+c(1除以a+1除以b)的值a+b+c=0,则a(1除以b+1除以c)+b(1除以a+1除以?

最佳回答

  • a/(1/b+1/c)+b/(1/a+1/c)+c/(1/a+1/b) =a/(c/bc+b/bc)+b/(c/ac+a/ac)+c/(b/ab+a/ab) =a/[(b+c)/bc]+b/[(a+c)/ac]+c/[(a+b)/ab] =abc/(b+c)+abc/(a+c)+abc/(a+b) =abc/(-a)+abc/(-b)+abc/(-c) =-bc-ac-ab =-a(b+c)-bc =-a*(-a)-bc =a^2-bc
    2012-04-24 21:18:34
  • 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+b)(a^2-ab+b^2)+c^3)/(abc) =-(c^2-a^2+ab-b^2)/(ab) =-(a^2+2ab+b^2-a^2+ab-b^2)/(ab) =-(3ab)/(ab) =-3
    2012-04-24 21:12:08
  • ∵a+b+c=0， ∴a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) =(a+b)/c+(b+c)/a+(c+a)/b =(-c)/c+(-a)/a+(-b)/b =-1-1-1 =-3。
    2012-04-24 20:56:51
  • 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 =-b/b+(-a/a)+(-c/c) =-3
    2012-04-24 20:47:05
  • 解：∵a+b+c=0,两边同时除以a 得：b/a+c/a=-1 同理：两边同时除以b得：a/b+c/b=-1; 两边同时除以c得：a/c+b/c=-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/a+c/a)+(a/b+c/b)+(a/c+b/c) =-1+(-1)+(-1) =-3。
    2012-04-24 20:45:15
    • 很赞哦！ (213)

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