Answer
$\dfrac{x^2+4}{x^4-8x^2+16}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{x\cdot2x-(x^2-4)\cdot1}{(x^2-4)^2}
,$ multiply the factors first. Then remove the grouping symbols and combine like terms. Finally, use special products to simplify the denominator.
$\bf{\text{Solution Details:}}$
Multiplying the factors, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2x^2-(x^2-4)}{(x^2-4)^2}
.\end{array}
Removing the grouping symbols and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2x^2-x^2+4}{(x^2-4)^2}
\\\\=
\dfrac{x^2+4}{(x^2-4)^2}
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{x^2+4}{(x^2)^2-2(x^2)(4)+(4)^2}
\\\\=
\dfrac{x^2+4}{x^4-8x^2+16}
.\end{array}
