Đặt \(x+2=a\)
\(\Rightarrow P=\left(a-1\right)^4+\left(a+1\right)^4\)
\(P=a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1\)
\(P=2a^4+12a^2+2\)
Do \(\left\{{}\begin{matrix}a^4\ge0\\a^2\ge0\end{matrix}\right.\) \(\forall a\Rightarrow P\ge0+0+2=2\)
\(\Rightarrow P_{min}=2\) khi \(a=0\Rightarrow x=-2\)
Đặt \(t=x+2\), ta được:
\(P=\left(t-1\right)^4+\left(t+1\right)^4\\ =2t^4+12t^2+2\\ =2t^2\left(t^2+6\right)+2\ge2\left(\forall t\in R\right)\)
Hay \(P\ge2\left(\forall x\in R\right)\)
Đẳng thức xảy ra\(\Leftrightarrow2t^2\left(t^2+6\right)=0\Leftrightarrow2t^2=0\Leftrightarrow t=0\Leftrightarrow x=-2\)
Vậy \(minP=2\), đạt được khi \(x=-2\)
