## Precalculus (6th Edition) Blitzer

$-\displaystyle \frac{2(x+2)}{x+1}\qquad x\neq 1,-1,2, -2$
Start by recognizing a difference of squares in the denominator of the denominator. $\displaystyle \frac{\frac{x}{x-2}+1}{\frac{3}{x^{2}-4}+1}= \frac{\frac{x}{x-2}+1}{\frac{3}{(x-2)(x+2)}-1}=...$ Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, multiplying both the numerator and denominator with $(x-2)(x+2)$ would get rid of these fractions. Currently, we note that we have to exclude $\pm 2$ as they yield zero in corresponding denominators.. The denominator of the complex rational expression must not be zero, so we will exclude any values for which it is 0 $\displaystyle \frac{3}{(x-2)(x+2)}-1=0\quad/\times (x-2)(x+2)$ $3-1(x^{2}-4)=0$ $3-x^{2}+4=0$ $1-x^{2}=0$ $(1-x)(1+x)=0$ $x=\pm 1$ must be also excluded. $\displaystyle \frac{\frac{x}{x-2}+1}{\frac{3}{(x-2)(x+2)}-1}\times\frac{(x-2)(x+2)}{(x-2)(x+2)}=\frac{x(x+2)+(x+2)(x-2)}{3-(x-2)(x+2)}$ ... the denominator was simplified while solving the equation, above. $=\displaystyle \frac{(x+2)[x+(x-2)]}{(1-x)(1+x)}$ $=\displaystyle \frac{(x+2)[2x-2]}{(1-x)(1+x)}$ $=\displaystyle \frac{(x+2)[2x-2]}{(1-x)(1+x)}\qquad$... factor out -2 from $2x-2$ $=-\displaystyle \frac{2(x+2)(1-x)}{(1-x)(x+1)}\qquad$... reduce the common factor $=-\displaystyle \frac{2(x+2)}{x+1}\qquad x\neq 1,-1,2, -2$