Answer
$x-a$
Work Step by Step
Expressing as similar fractions, the given expression, $
\dfrac{x-\dfrac{a^2}{x}}{1+\dfrac{a}{x}}
,$ is equivalent to
\begin{align*}
&
\dfrac{x\cdot\dfrac{x}{x}-\dfrac{a^2}{x}}{1\cdot\dfrac{x}{x}+\dfrac{a}{x}}
\\\\&=
\dfrac{\dfrac{x^2}{x}-\dfrac{a^2}{x}}{\dfrac{x}{x}+\dfrac{a}{x}}
\\\\&=
\dfrac{\dfrac{x^2-a^2}{x}}{\dfrac{x+a}{x}}
\\\\&=
\dfrac{x^2-a^2}{x}\div\dfrac{x+a}{x}
.\end{align*}
Multiplying by the reciprocal of the divisor and cancelling common factors, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{x^2-a^2}{x}\cdot\dfrac{x}{x+a}
\\\\&=
\dfrac{x^2-a^2}{\cancel x}\cdot\dfrac{\cancel x}{x+a}
\\\\&=
\dfrac{x^2-a^2}{x+a}
.\end{align*}
Using $a^2-b^2=(a+b)(a+b)$ or the factoring of the difference of two squares, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{(x+a)(x-a)}{x+a}
\\\\&=
\dfrac{(\cancel{x+a})(x-a)}{\cancel{x+a}}
&(\text{cancel common factor})
\\\\&=
x-a
.\end{align*}
Hence, the expression $\dfrac{x-\dfrac{a^2}{x}}{1+\dfrac{a}{x}}$ simplifies to $x-a$.
