2/(x-2)-K/(x²-4)=3/(x+2)
[2(x+2)-k-3(x-2)]/(x²-4)=0
2(x+2)-k-3(x-2)=0
x=-k+10
原方程增根只能为x=2或x=-2
-k+10=2或-k+10=-2
可得:k=8或k=12