Tìm GTNN của các biểu thức: M = x^2 + y^2 - x + 6y +10

* Hướng dẫn giải

$\begin{array}{l}\mathrm{M}={\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+6\mathrm{y}+10\\ =\left({\mathrm{x}}^2-\mathrm{x}\right)+\left({\mathrm{y}}^2+6\mathrm{y}\right)+10\\ =\left[\left({\mathrm{x}}^2-2\mathrm{x}.\frac{1}{2}+\frac{1}{4}\right)+\left({\mathrm{y}}^2+2\mathrm{y}.3+3^2\right)-\frac{1}{4}-9\right]+10\\ ={\left(\mathrm{x}-\frac{1}{2}\right)}^2+{\left(\mathrm{y}+3\right)}^2+\frac{3}{4}\\ \mathrm{Vì} {\left(\mathrm{x}-\frac{1}{2}\right)}^2\ge 0,{\left(\mathrm{y}+3\right)}^2\ge 0 (\mathrm{với} \mathrm{mọi} \mathrm{x}, \mathrm{y})\\ \Rightarrow \mathrm{M}\ge \frac{3}{4}. \mathrm{Dấu} \text{'}\text{'}=\text{'}\text{'} \mathrm{xảy} \mathrm{ra} \iff \left\{\begin{array}{l}\mathrm{x}-\frac{1}{2}=0\\ \mathrm{y}+3=0\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=\frac{1}{2}\\ \mathrm{y}=-3\end{array}\right.\\ \mathrm{Vậy} \mathrm{MinM}=\frac{3}{4}\iff \left\{\begin{array}{l}\mathrm{x}=\frac{1}{2}\\ \mathrm{y}=-3\end{array}\right.\end{array}$