Tìm GTNN: 3x^2+y^2+4x-7

Đáp án:

$P_{min}=-\dfrac{25}{3}$

Giải thích các bước giải:

$P=3x^2+y^2+4x-7$

     $=3\left(x^2+\dfrac{4}{3}x-\dfrac{7}{3}\right)+y^2$

     $=3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{25}{9}\right)+y^2$

     $=3\left[\left(x+\dfrac{2}{3}\right)^2-\dfrac{25}{9}\right]+y^2$

     $=3\left(x+\dfrac{2}{3}\right)^2-3.\dfrac{25}{9}+y^2$

     $=3\left(x+\dfrac{2}{3}\right)^2+y^2-\dfrac{25}{3}$

Ta có: $\begin{cases}3\left(x+\dfrac{2}{3}\right)^2\ge 0\\y^2\ge 0\end{cases}$

$⇒3\left(x+\dfrac{2}{3}\right)^2+y^2\ge 0$

$⇒3\left(x+\dfrac{2}{3}\right)^2+y^2-\dfrac{25}{3}\ge -\dfrac{25}{3}$

$⇒P\ge -\dfrac{25}{3}⇒P_{\min}=-\dfrac{25}{3}$

Dấu "=" xảy ra khi: $\begin{cases}3\left(x+\dfrac{2}{3}\right)^2=0\\y^2= 0\end{cases}⇒\begin{cases}x+\dfrac{2}{3}= 0\\y=0\end{cases}⇒\begin{cases}x=-\dfrac{2}{3}\\y=0\end{cases}$

Vậy $P_{min}=-\dfrac{25}{3}$ khi $(x;y)=\left(-\dfrac{2}{3};0\right)$.