Answer
$\displaystyle \frac{3x^{2}-4x+4}{(x-1)^{2}}$
Work Step by Step
Step 1:
Factor each denominator
$ x-1\qquad$ ... is factored,
Recognize $x^{2}\pm 2ax+a^{2}$ as a perfect square, $(x\pm a)^{2}$
$x^{2}-2x+1^{2}=(x-1)^{2}$
Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials.
LCM = $(x-1)^{2}$
Step 3:
Write each rational expression using the LCM as the denominator. Simplify.
$\displaystyle \frac{3x}{x-1}-\frac{x-4}{x^{2}-2x+1}=\frac{3x(x-1)}{(x-1)^{2}}-\frac{x-4}{(x-1)^{2}}$
$=\displaystyle \frac{3x^{2}-3x-x+4}{(x-1)^{2}}$
$=\displaystyle \frac{3x^{2}-4x+4}{(x-1)^{2}}$
For $ax^{2}+bx+c$ in the numerator,
we search for factors of ac whose sum is b, and rewrite the $bx$ term:
... no two factors of $+12$ add to $-4$, so we leave it as it is.
