a) \(A=5x^2-4x+1\)
\(=5\left(x^2-\frac{4}{5}x+\frac{1}{5}\right)\)
\(=5\left(x^2-\frac{4}{5}x+\frac{4}{25}-\frac{2}{25}\right)\)
\(=5\left[\left(x-\frac{2}{5}\right)^2-\frac{2}{25}\right]\)
\(=5\left[\left(x-\frac{2}{5}\right)^2\right]-2\ge-2\)
Vậy \(A_{min}=-2\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)
Sửa)):Dòng 3
\(=5\left(x^2-\frac{4}{5}x+\frac{4}{25}+\frac{1}{25}\right)\)
\(=5\left[\left(x-\frac{2}{5}\right)^2+\frac{1}{25}\right]\)
\(=5\left[\left(x-\frac{2}{5}\right)^2\right]+\frac{1}{5}\ge\frac{1}{5}\)
(Dấu "="\(\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)
