1/(x²-5x+6)-1/( x²-4x+3)+1/( x²-3x+2)=1/(x-1)
1/[(x-2)(x-3)]-1/[(x-1)(x-3)]+1/[(x-1)(x-2)]=1/(x-1)
1/(x-3)-1/（x-2）-1/[（x-1)(x-3）]+1/(x-2)-1/(x-1)=1/(x-1)
1/(x-3)-1/(x-1)-1/[(x-1)(x-3)]=1/(x-1)
2/[(x-3)(X-1)-1/[(x-1)(x-3)]=1/(x-1)
1/(x-3)(x-1)=1/(x-1)
x²-4x+3=x-1
x²-5x+4=0
（x-1)(x-4)=0
x=1（不合题意,舍去）
∴x=4