Answer
$\textbf{(a)}\dfrac{x+2}{x-2}$
$\textbf{(b)} \dfrac{x-1}{x-3}$
$\textbf{(c)} \dfrac{1}{x-2}$
$\textbf{(d)} -(y+x)$
Work Step by Step
$\textbf{(a)} \dfrac{x^2+3x+2}{x^2-x-2}=\dfrac{(x+1)(x+2)}{(x-2)(x+1)}=\dfrac{x+2}{x-2}$
$\textbf{(b)} \dfrac{2x^2-x-1}{x^2-9}.\dfrac{x+3}{2x+1}=\dfrac{(2x+1)(x-1)}{(x+3)(x-3)}.\dfrac{x+3}{2x+1}=\dfrac{x-1}{x-3}$
$\textbf{(c)} \dfrac{x^2}{x^2-4}-\dfrac{x+1}{x+2}=\dfrac{x^2}{(x-2)(x+2)}-\dfrac{x+1}{x+2}=\dfrac{x^2-(x+1)(x-2)}{(x-2)(x+2)}=\dfrac{x^2-(x^2-x-2)}{(x-2)(x+2)}=\dfrac{x+2}{(x-2)(x+2)}=\dfrac{1}{x-2}$
$\textbf{(d)} \dfrac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}=\dfrac{\frac{y^2-x^2}{xy}}{\frac{x-y}{yx}}=\dfrac{y^2-x^2}{x-y}=\dfrac{(y-x)(y+x)}{x-y}=-\dfrac{(y-x)(y+x)}{y-x}=-(y+x)$
