\(A=2x^2-6x+1\)
\(=2\left(x^2-3x+\frac{1}{2}\right)\)
\(=2\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{7}{4}\right]\)
\(=2\left[\left(x-\frac{3}{2}\right)^2+\frac{-7}{4}\right]\)
\(=2\left(x-\frac{3}{2}\right)^2+\frac{-7}{2}\)
Ta có : \(2\left(x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow2\left(x-\frac{3}{2}\right)^2+\frac{-7}{2}\ge\frac{-7}{2}\)
\(\Rightarrow A\ge\frac{-7}{2}\)
Dấu " = " xảy ra khi và chỉ khi \(x-\frac{3}{2}=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy \(Min_A=\frac{-7}{2}\Leftrightarrow x=\frac{3}{2}\)
Đúng 0
Bình luận (1)