若a+b+c=0,且abc≠0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+2的值

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若a+b+c=0,且abc≠0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+2的值
若a+b+c=0,且abc≠0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+2的值

若a+b+c=0,且abc≠0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+2的值
因a+b+c=0,所以a+b=-c;b+c=-a;a+c=-b;
则原式=a/b+a/c+b/a+b/c+c/a+c/b+2
=(a+c)/b+(a+b)/c+(b+c)/a+2
=(-b)/b+(-c)/c+(-a)/a+2
=-1+(-1)+(-1)+2
=-1