Ta có :
\(A=x^2-5x+8\)
\(=x^2-5x+\frac{25}{4}+\frac{7}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{7}{4}\)
Với mọi x ta có :
\(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
\(\Leftrightarrow A\ge\frac{7}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{5}{2}\)
Vậy...