Đặt \(a=x^2-4x-12\) thay vào N:
\(N=\left(x^2-4x-5\right)\left(x^2-4x-19\right)+49\)
\(=\left(a+7\right)\left(a-7\right)+49\)\(=a^2-49+49\)\(=a^2\)
Ta có: N = \(a^2\ge0\) \(\left(\forall a\right)\)
\(\Rightarrow\)MIN N = 0 \(\Leftrightarrow a^2=0\Leftrightarrow a=0\)
Hay \(x^2-4x-12=0\Leftrightarrow x^2-4x+4-16=0\)\(\Leftrightarrow\left(x-2\right)^2-4^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x=-2;x=6\)
Vậy Min A = 0 \(\Leftrightarrow x=-2;x=6\)