A=\(\left(x^2-4x-5\right)\left(x^2-4x-19\right)+49\)
Đặt \(x^2-4x-12=y\)
\(\Rightarrow\)A=\(\left(y+7\right)\left(y-7\right)+49\)
\(\Leftrightarrow\)A= \(y^2-49+49\)
\(\Leftrightarrow\)A=\(y^2\)
Ta có \(y^2\ge0\forall y\)
Hay \(\left(x^2-4x-12\right)^2\ge0\forall x\)
\(\Rightarrow\)Min A=0\(\Leftrightarrow\)\(x^2-4x-12=0\)
\(\Leftrightarrow\)\(\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+2=0\\x-6=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=6\end{cases}}\)