方程x-a-b/c+x-b-c/a+x-c-a/b=3,且1/a+1/b+1/c不等於0,x=??
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(x-a-b)/c-1+(x-b-c)/a-1+(x-c-a)/b-1=0(x-a-b-c)/c+(x-b-c-a)/a+(x-c-a-b)/b=0(x-a-b-c)(1/a+1/b+1/c)=0但1/a+1/b+1/c不等於0所以x-a-b-c=0x=a+b+c
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