\(A=-x^2+3x-3\)
\(=-\left(x^2-3x+3\right)\)
\(=-\left(x^2-3x+\frac{9}{4}\right)-\frac{3}{4}\)
\(=-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)
Dấu = khi \(x=\frac{3}{2}\)
\(-x^2+3x-3=-\left(x^2-3x+\frac{9}{4}\right)-\frac{3}{4}=-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\)
Vì: \(-\left(x-\frac{3}{2}\right)^2\le0\)
=> \(-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)
Vậy GTLN của bt trên là \(-\frac{3}{4}\) khi \(x=\frac{3}{2}\)
\(-x^2+3x-3=-\frac{107}{36}-x^2+2\times x\times\frac{1}{6}-\frac{1}{36}=-\frac{107}{36}-\left(x-\frac{1}{36}\right)^2\le-\frac{107}{36}\)
