当X取和值时,分式[X+3]除以{(X-2)(X+1)}有意义

答：k≠3
6/(x-1)=(x+k)/[x*(x-1)]-3/x
6x/[x*(x-1)]=(-2x^2+kx+3x)/[x*(x-1)]
6x=-2x^2+kx+3x
2x^2-kx+3x=0
x*(2x+3-k)=0
k≠3