Tìm GTLN: E= 3xmũ2 + 6x + 10 / xmũ2 + 2x + 3 15/10/2021 Bởi Eloise Tìm GTLN: E= 3xmũ2 + 6x + 10 / xmũ2 + 2x + 3
Đáp án: \(MaxE = \frac{7}{2}\) Giải thích các bước giải: \(\begin{array}{l}E = \frac{{3{x^2} + 6x + 10}}{{{x^2} + 2x + 3}}\\ = \frac{{3\left( {{x^2} + 2x + 3} \right) + 1}}{{{x^2} + 2x + 3}}\\ = 3 + \frac{1}{{{x^2} + 2x + 3}}\\ = 3 + \frac{1}{{{x^2} + 2x.1 + 1 + 2}}\\ = 3 + \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}}\\Do:{\left( {x + 1} \right)^2} \ge 0\forall x \in R\\ \to {\left( {x + 1} \right)^2} + 2 \ge 2\\ \to \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \le \frac{1}{2}\\ \to 3 + \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \le \frac{1}{2} + 3 = \frac{7}{2}\\ \to E \le \frac{7}{2}\\ \to MaxE = \frac{7}{2}\\ \Leftrightarrow x + 1 = 0\\ \to x = – 1\end{array}\) Bình luận
Đáp án:
\(MaxE = \frac{7}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
E = \frac{{3{x^2} + 6x + 10}}{{{x^2} + 2x + 3}}\\
= \frac{{3\left( {{x^2} + 2x + 3} \right) + 1}}{{{x^2} + 2x + 3}}\\
= 3 + \frac{1}{{{x^2} + 2x + 3}}\\
= 3 + \frac{1}{{{x^2} + 2x.1 + 1 + 2}}\\
= 3 + \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}}\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x + 1} \right)^2} + 2 \ge 2\\
\to \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \le \frac{1}{2}\\
\to 3 + \frac{1}{{{{\left( {x + 1} \right)}^2} + 2}} \le \frac{1}{2} + 3 = \frac{7}{2}\\
\to E \le \frac{7}{2}\\
\to MaxE = \frac{7}{2}\\
\Leftrightarrow x + 1 = 0\\
\to x = – 1
\end{array}\)