Ta có: \(\frac{x^2+4x+4}{x^2-4}=\frac{x^2+3x+2}{A}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+1\right)\left(x+2\right)}{A}\)
\(\Leftrightarrow\frac{x+2}{x-2}=\frac{\left(x+1\right)\left(x+2\right)}{A}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)\left(x+2\right)\left(x-2\right)}{x+2}\)
\(\Leftrightarrow A=\left(x+1\right)\left(x-2\right)=x^2-2x+x-2\)
hay \(A=x^2-x-2\)
Vậy: \(A=x^2-x-2\)