Tìm GTNN hoặc GTLN a. $\frac{1}{x^2+4}$ b. -8 * √(3-2x) c. x + $\frac{4}{x-1}$ khoảng ( 1 , dương vô cực) 29/08/2021 Bởi Remi Tìm GTNN hoặc GTLN a. $\frac{1}{x^2+4}$ b. -8 * √(3-2x) c. x + $\frac{4}{x-1}$ khoảng ( 1 , dương vô cực)
Đáp án: $\begin{array}{l}a)\dfrac{1}{{{x^2} + 4}}\\Do:{x^2} \ge 0\\ \Rightarrow {x^2} + 4 \ge 4\\ \Rightarrow \dfrac{1}{{{x^2} + 4}} \le \dfrac{1}{4}\\ \Rightarrow GTLN: = \dfrac{1}{4}\,khi:x = 0\\b) – 8.\sqrt {3 – 2x} \\Do:\sqrt {3 – 2x} \ge 0\\ \Rightarrow – 8.\sqrt {3 – 2x} \le 0\\ \Rightarrow GTLN = 0\,khi:x = \dfrac{3}{2}\\c)x + \dfrac{4}{{x – 1}}\left( {x > 1} \right)\\ = x – 1 + \dfrac{4}{{x – 1}} + 1\\Theo\,Co – si:\\\left( {x – 1} \right) + \dfrac{4}{{x – 1}} \ge 2.\sqrt {\left( {x – 1} \right).\dfrac{4}{{x – 1}}} = 4\\ \Rightarrow \left( {x – 1} \right) + \dfrac{4}{{x – 1}} + 1 \ge 4 + 1 = 5\\ \Rightarrow BT \ge 5\\ \Rightarrow GTNN: = 5\,khi:\left( {x – 1} \right) = \dfrac{4}{{x – 1}}\\ \Rightarrow {\left( {x – 1} \right)^2} = 4\\ \Rightarrow x – 1 = 2\\ \Rightarrow x = 3\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\dfrac{1}{{{x^2} + 4}}\\
Do:{x^2} \ge 0\\
\Rightarrow {x^2} + 4 \ge 4\\
\Rightarrow \dfrac{1}{{{x^2} + 4}} \le \dfrac{1}{4}\\
\Rightarrow GTLN: = \dfrac{1}{4}\,khi:x = 0\\
b) – 8.\sqrt {3 – 2x} \\
Do:\sqrt {3 – 2x} \ge 0\\
\Rightarrow – 8.\sqrt {3 – 2x} \le 0\\
\Rightarrow GTLN = 0\,khi:x = \dfrac{3}{2}\\
c)x + \dfrac{4}{{x – 1}}\left( {x > 1} \right)\\
= x – 1 + \dfrac{4}{{x – 1}} + 1\\
Theo\,Co – si:\\
\left( {x – 1} \right) + \dfrac{4}{{x – 1}} \ge 2.\sqrt {\left( {x – 1} \right).\dfrac{4}{{x – 1}}} = 4\\
\Rightarrow \left( {x – 1} \right) + \dfrac{4}{{x – 1}} + 1 \ge 4 + 1 = 5\\
\Rightarrow BT \ge 5\\
\Rightarrow GTNN: = 5\,khi:\left( {x – 1} \right) = \dfrac{4}{{x – 1}}\\
\Rightarrow {\left( {x – 1} \right)^2} = 4\\
\Rightarrow x – 1 = 2\\
\Rightarrow x = 3
\end{array}$