\(B=4x^2-x+10\)
\(\Leftrightarrow B=4x^2-2\times\dfrac{1}{4}\times2x+\dfrac{1}{16}+\dfrac{159}{16}\)
\(\Leftrightarrow B=\left(2x-\dfrac{1}{4}\right)^2+\dfrac{159}{16}\ge\dfrac{159}{16}\)(Do \(\left(2x-\dfrac{1}{4}\right)^2\ge0\))
Đẳng thức xảy ra khi: \(2x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{8}\)