Đáp án:
$M_{\min}=-\dfrac{85}{3}$
Giải thích các bước giải:
$M=3x^2-8x+y^2+4y-19$
$=3\left(x^2-2.x\dfrac{4}{3}+\dfrac{16}{9}\right)+(y^2+4y+4)-\dfrac{85}{3}$
$=3\left(x-\dfrac{4}{3}\right)^2+(y+2)^2-\dfrac{85}{3}$
Ta có: $\begin{cases}\left(x-\dfrac{4}{3}\right)^2\ge 0\\(y+2)^2\ge 0\end{cases}$
$⇒3\left(x-\dfrac{4}{3}\right)^2+(y+2)^2\ge 0$
$⇒3\left(x-\dfrac{4}{3}\right)^2+(y+2)^2-\dfrac{85}{3}\ge -\dfrac{85}{3}$
$⇒M\ge -\dfrac{85}{3}⇒M_{\min}=-\dfrac{85}{3}$
Dấu "=" xảy ra khi: $\begin{cases}\left(x-\dfrac{4}{3}\right)^2= 0\\(y+2)^2= 0\end{cases}⇒\begin{cases}x=\dfrac{4}{3}\\y=-2\end{cases}$
Vậy $M_{\min}=-\dfrac{85}{3}$ khi $(x;y)=\left(\dfrac{4}{3};-2\right)$.
