Question:
If $\frac{3x^2-4x+1}{x-1}=m$, and $x$ can be any real number except $1$, what real values can $m$ NOT have?

Answer:
We notice that the numerator of the fraction factors into $(3x-1)(x-1)$. Substituting this into the given expression, we get $m=\dfrac{3x^2-4x+1}{x-1} = \dfrac{(3x-1)(x-1)}{x-1}$. This simplifies to $m=3x-1$ if $x$ is not 1. Thus, $m$ can be any real number except for the value it takes when $x$ is $1.$ This value is $3(1)-1=3-1=\boxed{2}$.