a) ĐKXĐ : x $\neq$ 0
b) A = ($\frac{4}{x}$ - 4 - $\frac{4}{x}$ + 4) . $x^{2}$ + 8x + $\frac{16}{32}$
=[ $\frac{4}{x}$ ( 4 - 4 )]. $x^{2}$ + 8x + $\frac{1}{2}$
=8x + $\frac{1}{2}$
c) A = $\frac{1}{3}$
⇒8x + $\frac{1}{2}$ = $\frac{1}{3}$
⇔ x = -$\frac{1}{48}$ (TMĐKXĐ)
d) A = 8x + $\frac{1}{2}$ =