Đáp án:

c. \(MaxC = \dfrac{{145}}{4}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.A =  - 10{x^2} - 12x + 33\\
 =  - \left( {10{x^2} + 12x - 33} \right)\\
 =  - \left[ {{{\left( {x\sqrt {10} } \right)}^2} + 2.x\sqrt {10} .\dfrac{6}{{\sqrt {10} }} + {{\left( {\dfrac{6}{{\sqrt {10} }}} \right)}^2} - \dfrac{{183}}{5}} \right]\\
 =  - {\left( {x\sqrt {10}  + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5}\\
Do:{\left( {x\sqrt {10}  + \dfrac{6}{{\sqrt {10} }}} \right)^2} \ge 0\forall x\\
 \to  - {\left( {x\sqrt {10}  + \dfrac{6}{{\sqrt {10} }}} \right)^2} \le 0\\
 \to  - {\left( {x\sqrt {10}  + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5} \le \dfrac{{183}}{5}\\
 \to MaxA = \dfrac{{183}}{5}\\
 \Leftrightarrow x\sqrt {10}  + \dfrac{6}{{\sqrt {10} }} = 0\\
 \Leftrightarrow x =  - \dfrac{3}{5}\\
b.B =  - 9{x^2} - 12x + 5\\
 =  - \left( {9{x^2} + 12x - 5} \right)\\
 =  - \left( {9{x^2} + 2.3x.2 + 4 - 9} \right)\\
 =  - {\left( {3x + 2} \right)^2} + 9\\
Do:{\left( {3x + 2} \right)^2} \ge 0\\
 \to  - {\left( {3x + 2} \right)^2} \le 0\\
 \to  - {\left( {3x + 2} \right)^2} + 9 \le 9\\
 \to MaxB = 9\\
 \Leftrightarrow 3x + 2 = 0\\
 \Leftrightarrow x =  - \dfrac{2}{3}\\
c.C =  - {x^2} + 5x + 30\\
 =  - \left( {{x^2} - 5x - 30} \right)\\
 =  - \left( {{x^2} - 2x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{145}}{4}} \right)\\
 =  - {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4}\\
Do:{\left( {x - \dfrac{5}{2}} \right)^2} \ge 0\\
 \to  - {\left( {x - \dfrac{5}{2}} \right)^2} \le 0\\
 \to  - {\left( {x - \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4} \le \dfrac{{145}}{4}\\
 \to MaxC = \dfrac{{145}}{4}\\
 \Leftrightarrow x - \dfrac{5}{2} = 0\\
 \Leftrightarrow x = \dfrac{5}{2}
\end{array}\)

a,  

$-10x^2-12x+33$

$= -(10x^2+12x-33)$

$= -[(\sqrt{10}x)^2 - 2.\sqrt{10}x.\frac{3}{\sqrt{10}} + \frac{9}{10}-\frac{339}{10}]$

$= -(\sqrt{10}x^2-\frac{3}{\sqrt{10}})^2+\frac{339}{10}\le \frac{339}{10}$

$\Rightarrow max=\frac{339}{10}\Leftrightarrow x=\frac{3}{10}$

b,

$-9x^2-12x+5$

$= -(9x^2+12x-5)$

$= -[(3x)^2-2.3x.2 + 4 - 9]$

$= -(3x-2)^2+9\le 9$ 

$\Rightarrow max=9\Leftrightarrow x=\frac{2}{3}$ 

c, 

$-x^2+5x+30$

$=-(x^2-5x-30)$

$= -(x^2-2x.\frac{5}{2}+\frac{25}{4}-\frac{154}{4})$

$=-(x-\frac{5}{2})^2+\frac{154}{4}\le \frac{154}{4}$

$\Rightarrow max=\frac{154}{4}\Leftrightarrow x=\frac{5}{2}$