\(-A=2x^2-0,5x+8\)
=> \(-2A=4x^2-x+16\)
=> \(-2A=\left(2x-\frac{1}{4}\right)^2+\frac{255}{16}\)
Có: \(\left(2x-\frac{1}{4}\right)^2\ge0\)
=> \(-2A\ge\frac{255}{16}\)
=> \(A\le-\frac{255}{16}:2\)
=> \(A\le-\frac{255}{32}\)
DẤU "=" XẢY RA <=> \(\left(2x-\frac{1}{4}\right)^2=0\)
<=> \(x=\frac{1}{8}\)