Answer
$\dfrac{4}{x+2}$
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$, or the factoring of the difference of two squares, the given expression, $
\dfrac{1}{x-2}-\dfrac{4}{x^2-4}+\dfrac{3}{x+2}
,$ is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{1}{x-2}-\dfrac{4}{(x+2)(x-2)}+\dfrac{3}{x+2}
.\end{align*}
The $LCD$ of the denominators in the expression above is $(x+2)(x-2).$ Converting all terms into similar fractions, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{1}{x-2}\cdot\dfrac{x+2}{x+2}-\dfrac{4}{(x+2)(x-2)}+\dfrac{3}{x+2}\cdot\dfrac{x-2}{x-2}
\\\\&=
\dfrac{x+2}{(x+2)(x-2)}-\dfrac{4}{(x+2)(x-2)}+\dfrac{3(x-2)}{(x+2)(x-2)}
\\\\&=
\dfrac{x+2}{(x+2)(x-2)}-\dfrac{4}{(x+2)(x-2)}+\dfrac{3x-6}{(x+2)(x-2)}
.\end{align*}
Combining the numerators and copying the common denominator, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{x+2-4+3x-6}{(x+2)(x-2)}
\\\\&=
\dfrac{4x-8}{(x+2)(x-2)}
\\\\&=
\dfrac{4(x-2)}{(x+2)(x-2)}
&(\text{factor the common factor})
\\\\&=
\dfrac{4\cancel{(x-2)}}{(x+2)\cancel{(x-2)}}
&(\text{cancel the common factor})
\\\\&=
\dfrac{4}{x+2}
.\end{align*}
Hence, the expression $\dfrac{1}{x-2}-\dfrac{4}{x^2-4}+\dfrac{3}{x+2}$ is equivalent to $\dfrac{4}{x+2}$.
