Answer
The required solution $-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$.
Work Step by Step
We have the given complex rational expression
$\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}$.
We know that a rational expression is the one that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator, $q\ne 0$.
Furthermore, a complex rational expression or a complex fraction is an algebraic rational expression in which either the numerator or denominator or both the numerator and denominator contain rational expressions.
And the given complex rational expression is
$\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}$.
The numerator is $\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}$. Simplify the numerator:
$\begin{align}
& \frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}=\frac{1}{{{\left( x+h \right)}^{2}}}\times \frac{{{x}^{2}}}{{{x}^{2}}}-\frac{1}{{{x}^{2}}}\times \frac{{{\left( x+h \right)}^{2}}}{{{\left( x+h \right)}^{2}}} \\
& =\frac{{{x}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}-\frac{{{\left( x+h \right)}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\
& =\frac{{{x}^{2}}-{{\left( x+h \right)}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\
& =\frac{{{x}^{2}}-\left( {{x}^{2}}+2xh+{{h}^{2}} \right)}{{{x}^{2}}{{\left( x+h \right)}^{2}}}
\end{align}$.
Simplifying further,
$\begin{align}
& \frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}=\frac{{{x}^{2}}-{{x}^{2}}+2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\
& =\frac{2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}.
\end{align}$
And simplify the given complex rational expression:
$\begin{align}
& \frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}=\frac{\frac{-2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}}{h} \\
& =\frac{-h\left( 2x+h \right)}{{{x}^{2}}{{\left( x+h \right)}^{2}}}\cdot \frac{1}{h} \\
& =-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}
\end{align}$.
Hence, $\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}=$ $-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$ ; $x\ne 0\ ;\ h\ne 0\ ;\ x\ne h$.
