Answer
$\displaystyle \frac{x}{x+3},\qquad x\neq-3,-2$
Work Step by Step
Complex rational expressions have rational expressions in the numerator or/and in the denominator.
Here, the numerator contains $\displaystyle \frac{x}{x+3}$ (exclusion from the domain: $x\neq-3.\ )$
To get rid of $\displaystyle \frac{x}{x+3}$, we multiply both the numerator and denominator with $(x+3).$
$\displaystyle \frac{x+3}{x+3}\times\frac{x-\frac{x}{x+3}}{x+2}=\frac{x(x+3)-x}{(x+3)(x+2)},\qquad x\neq-3,-2$
$=\displaystyle \frac{x^{2}+3x-x}{(x+3)(x+2)}$
$=\displaystyle \frac{x^{2}+2x}{(x+3)(x+2)}$
$=\displaystyle \frac{x(x+2)}{(x+3)(x+2)}$
The expression has a common factor. Reduce.
$=\displaystyle \frac{x}{x+3},\qquad x\neq-3,-2$
