\(A=a^4-2a^3+3a^2-4a+5\)
\(\Leftrightarrow A=a^4-2a^3+a^2+2a^2-4a+2+3\)
\(\Leftrightarrow A=\left(a^4-2a^3+^2\right)+2\left(a^2-2a+1\right)+3\)
\(\Leftrightarrow A=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\)
Có:\(\hept{\begin{cases}\left(a^2-a\right)^2\ge0\forall x\\2\left(a-1\right)^2\ge0\forall x\end{cases}}\)
\(\Rightarrow A\ge3\). Dấu "=" \(\Leftrightarrow\hept{\begin{cases}a^2-a=0\\a-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a^2=a\\a=1\end{cases}}}\)
Vậy Min A=3 đạt được khi a=1
Nguồn: DORAEMON (lazi.vn)
