Tìm giá trị lớn nhất của biểu thức: a) -10x^2-12x+33 b) -9x^2 – 12x + 5 c) -x^2 + 5x +30 18/09/2021 Bởi Emery Tìm giá trị lớn nhất của biểu thức: a) -10x^2-12x+33 b) -9x^2 – 12x + 5 c) -x^2 + 5x +30
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, $-x^2+5x+30$ $=-(x^2-5x-30)$ $= -(x^2-2x.\frac{5}{2}+\frac{25}{4}-\frac{154}{4})$ $=-(x-\frac{5}{2})^2+\frac{154}{4}\le \frac{154}{4}$ $\Rightarrow max=\frac{154}{4}\Leftrightarrow x=\frac{5}{2}$ Bình luận
Đáp án: c. \(MaxC = \dfrac{{145}}{4}\) Giải thích các bước giải: \(\begin{array}{l}a.A = – 10{x^2} – 12x + 33\\ = – \left( {10{x^2} + 12x – 33} \right)\\ = – \left[ {{{\left( {x\sqrt {10} } \right)}^2} + 2.x\sqrt {10} .\dfrac{6}{{\sqrt {10} }} + {{\left( {\dfrac{6}{{\sqrt {10} }}} \right)}^2} – \dfrac{{183}}{5}} \right]\\ = – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5}\\Do:{\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} \ge 0\forall x\\ \to – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} \le 0\\ \to – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5} \le \dfrac{{183}}{5}\\ \to MaxA = \dfrac{{183}}{5}\\ \Leftrightarrow x\sqrt {10} + \dfrac{6}{{\sqrt {10} }} = 0\\ \Leftrightarrow x = – \dfrac{3}{5}\\b.B = – 9{x^2} – 12x + 5\\ = – \left( {9{x^2} + 12x – 5} \right)\\ = – \left( {9{x^2} + 2.3x.2 + 4 – 9} \right)\\ = – {\left( {3x + 2} \right)^2} + 9\\Do:{\left( {3x + 2} \right)^2} \ge 0\\ \to – {\left( {3x + 2} \right)^2} \le 0\\ \to – {\left( {3x + 2} \right)^2} + 9 \le 9\\ \to MaxB = 9\\ \Leftrightarrow 3x + 2 = 0\\ \Leftrightarrow x = – \dfrac{2}{3}\\c.C = – {x^2} + 5x + 30\\ = – \left( {{x^2} – 5x – 30} \right)\\ = – \left( {{x^2} – 2x.\dfrac{5}{2} + \dfrac{{25}}{4} – \dfrac{{145}}{4}} \right)\\ = – {\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4}\\Do:{\left( {x – \dfrac{5}{2}} \right)^2} \ge 0\\ \to – {\left( {x – \dfrac{5}{2}} \right)^2} \le 0\\ \to – {\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4} \le \dfrac{{145}}{4}\\ \to MaxC = \dfrac{{145}}{4}\\ \Leftrightarrow x – \dfrac{5}{2} = 0\\ \Leftrightarrow x = \dfrac{5}{2}\end{array}\) Bình luận
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,
$-x^2+5x+30$
$=-(x^2-5x-30)$
$= -(x^2-2x.\frac{5}{2}+\frac{25}{4}-\frac{154}{4})$
$=-(x-\frac{5}{2})^2+\frac{154}{4}\le \frac{154}{4}$
$\Rightarrow max=\frac{154}{4}\Leftrightarrow x=\frac{5}{2}$
Đáp án:
c. \(MaxC = \dfrac{{145}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = – 10{x^2} – 12x + 33\\
= – \left( {10{x^2} + 12x – 33} \right)\\
= – \left[ {{{\left( {x\sqrt {10} } \right)}^2} + 2.x\sqrt {10} .\dfrac{6}{{\sqrt {10} }} + {{\left( {\dfrac{6}{{\sqrt {10} }}} \right)}^2} – \dfrac{{183}}{5}} \right]\\
= – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5}\\
Do:{\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} \ge 0\forall x\\
\to – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} \le 0\\
\to – {\left( {x\sqrt {10} + \dfrac{6}{{\sqrt {10} }}} \right)^2} + \dfrac{{183}}{5} \le \dfrac{{183}}{5}\\
\to MaxA = \dfrac{{183}}{5}\\
\Leftrightarrow x\sqrt {10} + \dfrac{6}{{\sqrt {10} }} = 0\\
\Leftrightarrow x = – \dfrac{3}{5}\\
b.B = – 9{x^2} – 12x + 5\\
= – \left( {9{x^2} + 12x – 5} \right)\\
= – \left( {9{x^2} + 2.3x.2 + 4 – 9} \right)\\
= – {\left( {3x + 2} \right)^2} + 9\\
Do:{\left( {3x + 2} \right)^2} \ge 0\\
\to – {\left( {3x + 2} \right)^2} \le 0\\
\to – {\left( {3x + 2} \right)^2} + 9 \le 9\\
\to MaxB = 9\\
\Leftrightarrow 3x + 2 = 0\\
\Leftrightarrow x = – \dfrac{2}{3}\\
c.C = – {x^2} + 5x + 30\\
= – \left( {{x^2} – 5x – 30} \right)\\
= – \left( {{x^2} – 2x.\dfrac{5}{2} + \dfrac{{25}}{4} – \dfrac{{145}}{4}} \right)\\
= – {\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4}\\
Do:{\left( {x – \dfrac{5}{2}} \right)^2} \ge 0\\
\to – {\left( {x – \dfrac{5}{2}} \right)^2} \le 0\\
\to – {\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{145}}{4} \le \dfrac{{145}}{4}\\
\to MaxC = \dfrac{{145}}{4}\\
\Leftrightarrow x – \dfrac{5}{2} = 0\\
\Leftrightarrow x = \dfrac{5}{2}
\end{array}\)