ĐK: \(x\ne-4\)
\(A=\frac{x}{\left(x+4\right)^2}=\frac{16x}{16\left(x^2+8x+16\right)}=\frac{x^2+8x+16-x^2+8x-16}{16\left(x^2+8x+16\right)}=\frac{1}{16}-\frac{\left(x-4\right)^2}{16\left(x+4\right)^2}\le\frac{1}{16}\forall x\)
Dấu "=" xảy ra khi: \(x-4=0\Rightarrow x=4\) (thỏa mãn ĐKXĐ)
Vậy \(A_{max}=\frac{1}{16}\Leftrightarrow x=4\)