1 原方程可化为：1+1/(x+6)+1+1/(x+8)=1+1/(x+9)+1+1/(x+5) 既1/(x+6)-1/(x+5)=1/(x+9)-1/(x+8) 通分得-1/(x+6)(x+5)=-1/(x+9)(x+8) 既(x+6)(x+5)=(x+9)(x+8) 解得x=-7 2 方程右边同分得（-x+13）/（x-5)(x+3)=[(A-B)X+3A+5B]/(x-5)(x+3) 既-x+13=(A-B)x+3A+5B A-B=-1 3A+5B=13,A=1 B=2 3 a=1/bc b=1/ac a/(ab+a+1)=a/(1/c+a+1)=ac/(1+ac+c) b/(bc+b+1)=(1/ac)/(1/a+1/ac+1)=1/(ac+c+1) 故a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1) =ac/(1+ac+c)+1/(ac+c+1)+c/(ca+c+1) =(c+1+ac)/(ac+c+1)=1看看我的第一题的做法，比他们都简单............
