\(4x^2-x-\dfrac{3}{16}\)
\(=\left(4x^2-x+\dfrac{1}{16}\right)-\dfrac{4}{16}\)
\(=\left(2x-\dfrac{1}{4}\right)^2-\dfrac{4}{16}\)
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 !!
\(4x^2-x-\dfrac{3}{16}=4x^2-4x\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{3}{16}\)
\(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{4}\)
Có \(\left(2x-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(2x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
\(\Rightarrow min=\dfrac{-1}{4}\Leftrightarrow2x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{4}\)
AAAA!!!!!!Mk nhâm \(\left(2x-\dfrac{1}{4}\right)^2\ge0\Rightarrow min=\dfrac{-1}{4}\Leftrightarrow x=\dfrac{1}{8}\)
