Answer
$\dfrac{1}{x-1}$
Work Step by Step
Adding the numerators and copying the denominator, the given expression, $
\dfrac{x-5}{x^2-4x+3}+\dfrac{2}{x^2-4x+3}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{x-5+2}{x^2-4x+3}
\\\\=
\dfrac{x-3}{x^2-4x+3}
\\\\=
\dfrac{x-3}{(x-3)(x-1)}
\\\\=
\dfrac{\cancel{x-3}}{(\cancel{x-3})(x-1)}
\\\\=
\dfrac{1}{x-1}
.\end{array}
