## Precalculus (6th Edition) Blitzer

The required solution is $\frac{1}{\left( x+h+1 \right)\left( x+1 \right)}$.
We have the given complex rational expression: $\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}$. We know that a rational expression is 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 $\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}$. The numerator is $\frac{x+h}{x+h+1}-\frac{x}{x+1}$. Simplify the numerator: \begin{align} & \frac{x+h}{x+h+1}-\frac{x}{x+1}=\frac{x+h}{x+h+1}\times \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{x}{x+1}\times \frac{\left( x+h+1 \right)}{\left( x+h+1 \right)} \\ & =\frac{\left( x+h \right)\left( x+1 \right)}{\left( x+h+1 \right)\left( x+1 \right)}-\frac{x\left( x+h+1 \right)}{\left( x+1 \right)\left( x+h+1 \right)} \\ & =\frac{\left( x+h \right)\left( x+1 \right)-x\left( x+h+1 \right)}{\left( x+h+1 \right)\left( x+1 \right)} \\ & =\frac{{{x}^{2}}+x+xh+h-{{x}^{2}}-xh-x}{\left( x+h+1 \right)\left( x+1 \right)}. \end{align}. Simplifying further, $\frac{x+h}{x+h+1}-\frac{x}{x+1}=\frac{h}{\left( x+h+1 \right)\left( x+1 \right)}$. And simplify the given complex rational expression: \begin{align} & \frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}=\frac{\frac{h}{\left( x+h+1 \right)\left( x+1 \right)}}{h} \\ & =\frac{1}{\left( x+h+1 \right)\left( x+1 \right)} \end{align}. Thus, $\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}=$ $\frac{1}{\left( x+h+1 \right)\left( x+1 \right)}$ ; $x\ne -1\ ;\ h\ne -\left( x+1 \right)$.