\(A=x^2-5x+7\)
\(=x^2-5x+\dfrac{25}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x ta có :
\(\left(x-\dfrac{5}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow A>0\)
Vậy..