Đáp án:

`M=-3x^2-x+5`

`=-3(x^2+1/3x)+5`

`=-3(x^2+1/3x+1/36-1/36)+5`

`=-3(x+1/6)^2+1/12+5`

`=-3(x+1/6)^2+61/12<=61/12`

Dấu "=" xảy ra khi `x+1/6=0<=>x=-1/6`

`N=3x+x^2-2`

`=x^2+3x-2`

`=x^2+3x+9/4-9/4-2`

`=(x+3/2)^2-17/4>=-17/4`

Dấu "=" xảy ra khi `x=-3/2`

`H=3x^2-y+2y^2+x-12`

`=3(x^2+1/3x)+2(y^2-1/2y)-12`

`=3(x^2+1/3x+1/36-1/36)+2(y^2-1/2y+1/16-1/16)-12`

`=3(x+1/6)^2+2(y-1/4)^2-1/12-1/8-12`

`=3(x+1/6)^2+2(y-1/4)^2-293/24>=-293/24`

Dấu "=" xảy ra khi \(\begin{cases}x+\dfrac16=0\\y-\dfrac14=0\\\end{cases}\)

`<=>` \(\begin{cases}x=-\dfrac16\\y=\dfrac14\\\end{cases}\)

Đáp án:

Không tồn tại x để M đạt GTNN

\(MinN =  - \dfrac{{17}}{4}\)

\(MinH =  - \dfrac{{293}}{{24}}\)

Giải thích các bước giải:

\(\begin{array}{l}
M =  - 3{x^2} - x + 5\\
 =  - \left( {3{x^2} + x - 5} \right)\\
 =  - \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}} - \dfrac{{61}}{{12}}} \right)\\
 =  - {\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}}\\
Do:{\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \ge 0\forall x\\
 \to  - {\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \le 0\\
 \to  - {\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}} \le \dfrac{{61}}{{12}}\\
 \to Max = \dfrac{{61}}{{12}}\\
 \Leftrightarrow x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }} = 0\\
 \to x =  - \dfrac{1}{6}
\end{array}\)

⇒ Không tồn tại x để M đạt GTNN

\(\begin{array}{l}
N = {x^2} + 3x - 2\\
 = {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{{17}}{4}\\
 = {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{4}\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0\forall x\\
 \to {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{4} \ge  - \dfrac{{17}}{4}\\
 \to Min =  - \dfrac{{17}}{4}\\
 \Leftrightarrow x =  - \dfrac{3}{2}\\
H = 3{x^2} - y + 2{y^2} + x - 12\\
 = 3{x^2} + x + 2{y^2} - y - 12\\
 = \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}}} \right) + \left( {2{y^2} - 2.y\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8}} \right) - \dfrac{{293}}{{24}}\\
 = {\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2  - \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{{293}}{{24}}\\
Do:{\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2  - \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x;y\\
 \to {\left( {x\sqrt 3  + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2  - \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{{293}}{{24}} \ge  - \dfrac{{293}}{{24}}\\
 \to Min =  - \dfrac{{293}}{{24}}\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - \dfrac{1}{6}\\
y =  - \dfrac{1}{4}
\end{array} \right.
\end{array}\)