\(16x^2-8xy+y^2+1=\left(4x-y\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow4x=y\)
\(-4x^2+2x-1=-\left(4x^2-2\cdot2\cdot\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{15}{16}=-\left(2x-\dfrac{1}{4}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\)
Dấu \("="\Leftrightarrow2x=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{8}\)