\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)
\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1\right)+2093\)
\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1+3\right)+2093\)
Đặt: \(a=x^2+4x+1\)
\(\Rightarrow P=a^2-12\left(a+3\right)+2093\)
\(P=a^2-12a-36+2093\)
\(P=a^2-12a+2057\)
\(P=a^2-12a+36+2021\)
\(P=\left(a^2-2\cdot6\cdot a+6^2\right)+2021\)
\(P=\left(a-6\right)^2+2021\)
Ta có: \(\left(a-6\right)^2\ge0\forall a\)
\(\Rightarrow P=\left(t-6\right)^2+2021\ge2021\)
\(\Rightarrow P\ge2021\Rightarrow P_{min}=2021\)
Dấu "=" xảy ra: \(\left(t-6\right)^2=0\Leftrightarrow t-6=0\Leftrightarrow t=6\)
Vậy: \(P_{min}=2021\) khi \(t=6\)
Mà: \(t=6\Rightarrow x^2+4x+1=6\)
\(\Leftrightarrow x^2+4x+1-6=0\)
\(\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy: \(P_{min}=2021\) khi \(\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\\ P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+4\right)+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6-36+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6+36+2021\\ P=\left(x^2+4x-5\right)^2+2021\ge2021\)
Dấu "=" xảy ra tương đương với \(\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)