$M=-3x^2-x+5\\=-3(x^2+\dfrac{1}{3}x-\dfrac{5}{3})\\=-3(x^2+2.x.\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{61}{36})\\=\dfrac{61}{12}-3(x+\dfrac{1}{6})^2$
Vì $(x+\dfrac{1}{6})^2\ge0\ \forall x\in R$ nên
$M\le\dfrac{61}{12}$
Dấu $=$ xảy ra khi
$x+\dfrac{1}{6}=0$ hay $x=\dfrac{-1}{6}$
Vậy $M_{max}=\dfrac{61}{12}$ khi $x=\dfrac{-1}{6}$
Tương tự như vậy, ta có
$N=3x+x^2-2\\=x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-2-\dfrac{9}{4}\\=(x+\dfrac{3}{2})^2-\dfrac{17}{4}\ge\dfrac{-17}{4}$
Dấu $=$ xảy ra khi
$x+\dfrac{3}{2}=9$ hay $x=\dfrac{-3}{2}$
Vậy $N_{min}=\dfrac{-17}{4}$ khi $x=\dfrac{-3}{2}$
$H=3x^2-y+2y^2+x-12\\=3(x^2+\dfrac{1}{3}x+\dfrac{1}{36})+2(y^2-\dfrac{1}{2}y+\dfrac{1}{16})-12-\dfrac{1}{12}-\dfrac{1}{8}\\=3(x+\dfrac{1}{6})^2+2(y-\dfrac{1}{4})^2-\dfrac{293}{24}\le\dfrac{-293}{24}$
Dấu $=$ xảy ra khi
$x+\dfrac{1}{6}=0\ ;\ y-\dfrac{1}{4}=0$ hay
$x=\dfrac{-1}{6}\ ;\ y=\dfrac{1}{4}$
Vậy $H_{min}=\dfrac{-293}{24}$ khi $(x;y)=\bigg(\dfrac{-1}{6};\dfrac{1}{4}\bigg)$
