a) Ta có:
\(M=4x^2-2x+1\)
\(=\left(2x\right)^2-2x.2.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(2x\right)^2-2x.2.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta lại có: \(\left(2x-\dfrac{1}{2}\right)^2\ge0\)
\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow M\ge\dfrac{3}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow\left(2x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(Min_M=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(M=4x^2-2x+1=\left(4x^2-2x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của M là \(\dfrac{3}{4}\) khi x = \(\dfrac{1}{4}\)
\(N=-x^2+x-2=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}\)
Vậy GTLN của N là \(-\dfrac{7}{4}\) khi x = \(\dfrac{1}{2}\)