A=2(x^2-2.5/4x+25/16)-50/16+7
A=2(x-√10/5)^2+31/8
Vì(x-√10/5)^2>=0 với mọi x
=>A>=31/8
Chọn B
\(2x^2-5x+7=2\left(x^2-\dfrac{5}{2}x+\dfrac{25}{16}\right)-\dfrac{25}{8}+7=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}+7\ge\dfrac{31}{8}\)
ĐTXR⇔\(x=\dfrac{5}{4}\)
Vậy minA =\(\dfrac{31}{8}\)khi x=\(\dfrac{5}{4}\)
Đáp án: \(B:\dfrac{31}{8}\)