\(a,=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)
Dấu \("="\Leftrightarrow x=3\)
\(b,=\left(x^2-2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{33}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\)
Dấu \("="\Leftrightarrow x=\dfrac{5}{2}\)
a) \(6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
Vậy ...
b) \(=\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{33}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
