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Câu hỏi: Giá trị nhỏ nhất của hàm số $y=3x+\dfrac{4}{{{x}^{2}}}$ trên khoảng $\left( 0;+\infty \right)$ là
A. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=3\sqrt[3]{9}.$
B. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=7.$
C. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=\dfrac{33}{5}.$
D. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=2\sqrt[3]{9}.$
A. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=3\sqrt[3]{9}.$
B. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=7.$
C. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=\dfrac{33}{5}.$
D. $\underset{\left( 0;+\infty \right)}{\mathop{\min }} y=2\sqrt[3]{9}.$
${y}'=3-\dfrac{8}{{{x}^{3}}};{y}'=0\Leftrightarrow 3-\dfrac{8}{{{x}^{3}}}=0\Leftrightarrow {{x}^{3}}=\dfrac{8}{3}\Leftrightarrow x=\dfrac{2}{\sqrt[3]{3}}$
$\Rightarrow y=3.\dfrac{2}{\sqrt[3]{3}}+\dfrac{4}{{{\left( \dfrac{2}{\sqrt[3]{3}} \right)}^{2}}}=3\sqrt[3]{9}$
$\Rightarrow y=3.\dfrac{2}{\sqrt[3]{3}}+\dfrac{4}{{{\left( \dfrac{2}{\sqrt[3]{3}} \right)}^{2}}}=3\sqrt[3]{9}$
Đáp án A.