(3/x-3)-(2/x-2)=(1/x-1)
[3(x-2)-2(x-3)]/(x-3)(x-2)=1/(x-1)
x/(x²-5x+6)=1/(x-1)
x(x-1)=x²-5x+6
5x-x=6
x=1.5
(3/x-3)-(2/x-2)=(1/x-1)是不是用十字相乘
(3/x-3)-(2/x-2)=(1/x-1)
[3(x-2)-2(x-3)]/(x-3)(x-2)=1/(x-1)
x/(x²-5x+6)=1/(x-1)
x(x-1)=x²-5x+6
5x-x=6
x=1.5