1)
x/(x-2)-(1-x²)/(x²-5x+6) = 2x/(x-3)
通分得
x(x-3)/[(x-2)(x-3)]-(1-x²)/[(x-2)(x-3)]=2x(x-2)/[(x-2)(x-3)]
x(x-3)-(1-x²)=2x(x-2)
x²-3x-1+x²=2x²-4x
x=1,且(x-2),(x-3)都非0,是原方程的解
2)
x-2分之2+(x^2-4)分之mx=0
2/(x-2)+mx/(x²-4)=0
(2(x+2)+mx)/(x²-4)=0
(2+m)x+4=0
当x=2或是-x时(x²-4)为0,
当x=2
2+m=-2
m=-4
当x=-2时
2+m=2
m=0
所以,m=0或是-4时方程会产生增根