1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0

第一题由X^2+3X+1=0,得X^2=-3X-1代入,1/(X+1)(X+2)=1,故原式=1/X(X+1)+1+1/(X+2)(X+3)=1+1/(-1-2X）+1/(2X+5)通分,原式=1+4/-(2X+5)(2X+1)=1+4/-[4(X^2+3X+1)+1]=1-4=-3 第二题移项,1/(x-10)-1/(x-7)=1/(x-9)-1/(x-6...