Question

tìm min \(M=x^4-2x^3+3x^2-4x+2025\)

Answer 1

\(M=x^4-2x^3+3x^2-4x+2025\\=(x^4-2x^3+x^2)+(2x^2-4x+2)+2023\\=x^2(x^2-2x+1)+2(x^2-2x+1)+2023\\=(x^2-2x+1)(x^2+2)+2023\\=(x-1)^2(x^2+2)+2023\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\x^2+2\ge2>0\forall x\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2\left(x^2+2\right)\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2\left(x^2+2\right)+2023\ge2023\forall x\)

\(\Rightarrow M\ge2023\forall x\) 

Dấu \("="\) xảy ra khi: \(x-1=0\Leftrightarrow x=1\)

Vậy \(Min_M=2023\) khi \(x=1\).