Tìm GTLN của biểu thức: B = x - x^2

* Hướng dẫn giải

$\begin{array}{l}\mathrm{B}=\mathrm{x}-{\mathrm{x}}^2=-\left({\mathrm{x}}^2-\mathrm{x}\right)=-\left[\left({\mathrm{x}}^2-2\mathrm{x}.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}\right]=-{\left(\mathrm{x}-\frac{1}{2}\right)}^2+\frac{1}{4}\\ \mathrm{Vì} {\left(\mathrm{x}-\frac{1}{2}\right)}^2\ge 0 (\mathrm{với} \mathrm{mọi} \mathrm{x})\Rightarrow -{\left(\mathrm{x}-\frac{1}{2}\right)}^2\le 0 (\mathrm{với} \mathrm{mọi} \mathrm{x})\\ \mathrm{Dấu} \text{'}\text{'}=\text{'}\text{'} \mathrm{xảy} \mathrm{ra} \iff \mathrm{x}-\frac{1}{2}=0\iff \mathrm{x}=\frac{1}{2}. \mathrm{Vậy} \mathrm{Max}\mathrm{B}=\frac{1}{4}\iff \mathrm{x}=\frac{1}{2}\\ \mathrm{R}=2\mathrm{x}-2{\mathrm{x}}^2-5=-2\left({\mathrm{x}}^2-\mathrm{x}+\frac{5}{2}\right)=-2\left[\left({\mathrm{x}}^2-2\mathrm{x}.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+\frac{5}{2}\right]\\ =-2{\left(\mathrm{x}-\frac{1}{2}\right)}^2-\frac{9}{2}\\ \mathrm{Vì} {\left(\mathrm{x}-\frac{1}{2}\right)}^2\ge 0 (\mathrm{với} \mathrm{mọi} \mathrm{x})\Rightarrow -2{\left(\mathrm{x}-\frac{1}{2}\right)}^2\le 0 (\mathrm{với} \mathrm{mọi} \mathrm{x})\\ \Rightarrow -2{\left(\mathrm{x}-\frac{1}{2}\right)}^2-\frac{9}{2}\le -\frac{9}{2}\left(\forall \mathrm{x}\right). \mathrm{Khi} \mathrm{đó}: \mathrm{x}-\frac{1}{2}=0\iff \mathrm{x}=\frac{1}{2}\\ \mathrm{Vậy} \mathrm{MaxR}=\frac{-9}{2}\iff \mathrm{x}=\frac{1}{2}\end{array}$