\(A=x^4-x^2+2x+2020\)
\(A=\left(x^4-2x^2+1\right)+\left(x^2+2x+1\right)+2018\)
\(A=\left(x^2-1\right)^2+\left(x+1\right)^2+2018\)
\(A=\left(x-1\right)^2\left(x+1\right)^2+\left(x+1\right)^2+2018\)
\(A=\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]+2018\)
Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left[\left(x-1\right)^2+1\right]>0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]\ge0\)
\(\Rightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]+2018\ge2018\)
\(\Rightarrow Amin=2018\Leftrightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x-1\right)^2+1=0\end{matrix}\right.\)
Mà \(\left(x-1\right)^2+1>0\) với mọi x
=> \(\left(x-1\right)^2+1\) vô nghiệm
\(\Rightarrow\left(x+1\right)^2=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
Vậy GTNN của A là 2018 khi x = -1
Vậy GTNN của A là 2018 khi x
