Có: y = $\frac{x-3}{x + 4}$
=> y' = $\frac{(x-3)'(x+4)-(x-3)(x+4)'}{(x + 4)²}$
= $\frac{((x+4)-(x-3)}{(x + 4)²}$
= $\frac{x+4-x+3}{(x + 4)²}$
= $\frac{1}{(x + 4)²}$
=> y'' = -$\frac{[(x+4)²]'}{(x+4)^{4}}$
= -$\frac{2(x+4)(x+4)'}{(x+4)^{4}}$
= -$\frac{2(x+4)}{(x+4)^{4}}$
= -$\frac{2}{(x+4)³}$
=> A = 2.($\frac{1}{(x + 4)²}$)² + (1 - $\frac{x-3}{x + 4}$).-$\frac{2}{(x+4)³}$
= $\frac{2}{(x + 4)^{4}}$ + $\frac{x+4-(x-3)}{x + 4}$.$\frac{-2}{(x+4)³}$
= $\frac{2}{(x + 4)^{4}}$ + $\frac{-14}{(x+4)^{4}}$
= $\frac{-12}{(x+4)^{4}}$
Vậy A = $\frac{-12}{(x+4)^{4}}$.
