Answer
$\displaystyle \frac{x}{(x-1)(x-2)}=\frac{-1}{x-1}+\frac{2}{x-2} $
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2} ~~~~~(1)$$
Now, write the RHS with a common denominator..
$$\displaystyle \frac{x}{(x-1)(x-2)}=\dfrac{(x-2)A+(x-1)B}{(x-1)(x-2)}\\ \displaystyle \frac{x}{(x-1)(x-2)}= \dfrac{Ax-2A+Bx-B}{(x-1)(x-2)}\\ \displaystyle \frac{x}{(x-1)(x-2)}=\dfrac{(A+B)x+(-2A-B)}{(x-1)(x-2)}$$
Next, equate the numerators to obtain:
$$(A+B)x+(-2A-B)=x$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+B=1 ~~~(2) \\ -2A-B=0 ~~~(3)$$
Equation (2) yields: $$A=1-B ~~~(4)$$
Plug $A=1-B$ into the Equation-(3) to solve for $B$.
Thus, we have: $$-2A-B=0\\ -2(1-B)-B=0 \\ -2+2B-B=0 \\-2+B=0\\ B=2$$
Finally, plug $B=2$ into equation (4) to obtain: and $A=1-2=-1$
Therefore, the equation (1) becomes: $\displaystyle \frac{x}{(x-1)(x-2)}=\frac{-1}{x-1}+\frac{2}{x-2} $
