\(A=x^4+6x^3+13x^2+12x+12\)
\(=\left(x^4+6x^3+19x^2+30x+25\right)-6x^2-18x-30+17\)
\(=\left(x^4+6x^3+19x^2+30x+25\right)-6\left(x^2+3x+5\right)+17\)
\(=\left(x^2+3x+5\right)^2-6\left(x^2+3x+5\right)+17\)
Đặt \(t=x^2+3x+5\)
Khi đó \(A=t^2-6t+17=t^2-2.t.3+9+8=\left(t-3\right)^2+8\ge8\)
Dấu "=" xảy ra <=> t - 3 = 0 <=> t = 3
<=> \(x^2+3x+5=3\Leftrightarrow x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy AMin = 8 khi và chỉ khi x = -1 hoặc x = -2
