Answer
$-\dfrac{8}{x}$
Work Step by Step
The factored form of the given expression, $
\dfrac{3}{2-x}-\dfrac{5}{x}+\dfrac{6}{x^2-2x}
,$ is
\begin{align*}
&
\dfrac{3}{2-x}-\dfrac{5}{x}+\dfrac{6}{x(x-2)}
\\\\&=
\dfrac{3}{-(-2+x)}-\dfrac{5}{x}+\dfrac{6}{x(x-2)}
\\\\&=
-\dfrac{3}{x-2}-\dfrac{5}{x}+\dfrac{6}{x(x-2)}
.\end{align*}
Changing each term to similar fractions by making the denominators equal to the $LCD=
x(x-2)
$, the expression above is equivalent to
\begin{align*}\require{cancel}
&
-\dfrac{3}{x-2}\cdot\dfrac{x}{x}-\dfrac{5}{x}\cdot\dfrac{x-2}{x-2}+\dfrac{6}{x(x-2)}
\\\\&=
-\dfrac{3x}{x(x-2)}-\dfrac{5(x-2)}{x(x-2)}+\dfrac{6}{x(x-2)}
\\\\&=
\dfrac{-3x-5(x-2)+6}{x(x-2)}
\\\\&=
\dfrac{-3x-5x+10+6}{x(x-2)}
\\\\&=
\dfrac{-8x+16}{x(x-2)}
\\\\&=
\dfrac{-8(x-2)}{x(x-2)}
\\\\&=
\dfrac{-8\cancel{(x-2)}}{x\cancel{(x-2)}}
\\\\&=
\dfrac{-8}{x}
\\\\&=
-\dfrac{8}{x}
.\end{align*}
Hence, the expression $
\dfrac{3}{2-x}-\dfrac{5}{x}+\dfrac{6}{x^2-2x}
$ simplifies to $
-\dfrac{8}{x}
$.
