a) ĐKXĐ: $x\ne 2;x\ne -3$
Khi đó: $A=\frac{x+2}{x+3}-\frac{5}{{{x}^{2}}+x-6}+\frac{1}{2-x}$
$A=\frac{x+2}{x+3}-\frac{5}{\left( x+3 \right)\left( x-2 \right)}-\frac{1}{x-2}$
$A=\frac{\left( x+2 \right)\left( x-2 \right)-5-\left( x+3 \right)}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{{{x}^{2}}-4-5-x-3}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{{{x}^{2}}-x-12}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{{{x}^{2}}-4x+3x-12}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{x\left( x-4 \right)+3\left( x-4 \right)}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{\left( x+3 \right)\left( x-4 \right)}{\left( x+3 \right)\left( x-2 \right)}$
$A=\frac{x-4}{x-2}$
b) Để $A=\frac{-3}{4}$ thì $\frac{x-4}{x-2}=\frac{-3}{4}$
$\Rightarrow 4\left( x-4 \right)=-3\left( x-2 \right)$
$\Leftrightarrow 4x-16=-3x+6$
$\Leftrightarrow 4x+3x=6+16$
$\Leftrightarrow 7x=22$
$\Leftrightarrow x=\frac{22}{7}$
Vậy với $x=\frac{22}{7}$ thì $A=\frac{-3}{4}$
d) Ta có: ${{x}^{2}}-9=0\Leftrightarrow {{x}^{2}}=9\Leftrightarrow \left[ \begin{align}
& x=3\,(tm) \\
& x=-3\,(ktm) \\
\end{align} \right.$
Thay x = 3 vào A ta được $A=\frac{3-4}{3-2}=-1$
Vậy với ${{x}^{2}}-9=0$ thì A = -1