Tim GTLN A= -X^2 + 6X -4 B=2018/X^2 + 2X +6 can gap a 20/11/2021 Bởi Parker Tim GTLN A= -X^2 + 6X -4 B=2018/X^2 + 2X +6 can gap a
Đáp án: $\begin{array}{l}A = – {x^2} + 6x – 4\\ = – \left( {{x^2} – 6x + 9} \right) + 9 – 4\\ = – {\left( {x – 3} \right)^2} + 5\\Do:{\left( {x – 3} \right)^2} \ge 0\forall x\\ \Rightarrow – {\left( {x – 3} \right)^2} \le 0\forall x\\ \Rightarrow – {\left( {x – 3} \right)^2} + 5 \le 5\forall x\\ \Rightarrow GTLN:A = 5 \Leftrightarrow x = 3\\B = \frac{{2018}}{{{x^2} + 2x + 6}}\\Do:{x^2} + 2x + 6\\ = {x^2} + 2x + 1 + 5\\ = {\left( {x + 1} \right)^2} + 5 \ge 5\forall x\\ \Rightarrow \frac{1}{{{x^2} + 2x + 6}} \le \frac{1}{5}\\ \Rightarrow \frac{{2018}}{{{x^2} + 2x + 6}} \le \frac{{2018}}{5}\\ \Rightarrow GTLN:B = \frac{{2018}}{5} \Leftrightarrow x = – 1\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = – {x^2} + 6x – 4\\
= – \left( {{x^2} – 6x + 9} \right) + 9 – 4\\
= – {\left( {x – 3} \right)^2} + 5\\
Do:{\left( {x – 3} \right)^2} \ge 0\forall x\\
\Rightarrow – {\left( {x – 3} \right)^2} \le 0\forall x\\
\Rightarrow – {\left( {x – 3} \right)^2} + 5 \le 5\forall x\\
\Rightarrow GTLN:A = 5 \Leftrightarrow x = 3\\
B = \frac{{2018}}{{{x^2} + 2x + 6}}\\
Do:{x^2} + 2x + 6\\
= {x^2} + 2x + 1 + 5\\
= {\left( {x + 1} \right)^2} + 5 \ge 5\forall x\\
\Rightarrow \frac{1}{{{x^2} + 2x + 6}} \le \frac{1}{5}\\
\Rightarrow \frac{{2018}}{{{x^2} + 2x + 6}} \le \frac{{2018}}{5}\\
\Rightarrow GTLN:B = \frac{{2018}}{5} \Leftrightarrow x = – 1
\end{array}$