分式的乘除：若x等于它的倒数,则[(x²-3x)/(x-3)]÷[(x-3)/(x²-6x+9)]
∵x=1/x,∴x²=1,x=±1；
[(x²-3x)/(x-3)]÷[(x-3)/(x²-6x+9)]=[x(x-3)/(x-3)]×[(x-3)²/(x-3)]=x(x-3)=x²-3x=1±3=4或-2.