\(4x^2-x-\dfrac{3}{16}\)
\(=\left(4x^2-x+\dfrac{1}{16}\right)-\dfrac{4}{16}\)
\(=\left(2x-\dfrac{1}{4}\right)^2-\dfrac{4}{16}\)
Do : \(\left(2x-\dfrac{1}{4}\right)^2\ge0\Rightarrow\left(2x-\dfrac{1}{4}\right)^2-\dfrac{4}{16}\ge-\dfrac{4}{16}=-\dfrac{1}{4}\)
Vậy GTNN của biểu thức là \(-\dfrac{1}{4}\) . Dấu \("="\) xảy ra khi \(\left(2x-\dfrac{1}{4}\right)^2=0\Leftrightarrow x=\dfrac{1}{8}\)
Wish you study well !!