Em $A$, temos: I) $\Large{\frac{x+2}{2x-3}}$ $ < 4$ e II) $\Large{\frac{x+2}{2x-3}}$ $ > -\space 4$ . I) $\Large{\frac{x+2}{2x-3}}$ $ -\space 4$ $ < 0$ $\implies$ $\Large{\frac{14-7x}{2x-3}}$ $ < 0$ $\implies$ $x>2$ ou $x<\large{\frac{3}{2}}$ . II) $\Large{\frac{x+2}{2x-3}}$ $ +\space 4$ $ > 0$ $\implies$ $\Large{\frac{9x-10}{2x-3}}$ $ > 0$ $\implies$ $x>\large{\frac{3}{2}}$ ou $x<\large{\frac{10}{9}}$ . Calculando a união entre I) e II), temos o intervalo $A =$ $]-\infty,\large{\frac{10}{9}}$$[\cup ]2, +\infty[$ . Em $B$, temos: $\log_{9} (x^2-5x+7)>0$ $\implies$ $x^2-5x+7>1$ , assim: $\Large{\left(x-\frac{5}{2} \right)^2}$ $> -6 + \Large{\frac{25}{4}} = \frac{1}{4}$ $\implies$ $\Large{\left|x-\frac{5}{2} \right|}$ $> \Large{\frac{1}{2}}$ , assim: $B =$ $]-\infty,2[\cup ]3, +\infty[$ . Portanto, temos que $A \cap B$ $=$ $]-\infty,\large{\frac{10}{9}}$$[\cup ]3, +\infty[$ Alternativa $\mathbb{(D)}$