Đáp án:
a) \(Min = \sqrt 6 - \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)y = \dfrac{{3x}}{2} + \dfrac{1}{{x + 1}}\left( {x > - 1} \right)\\
= \dfrac{{3\left( {x + 1} \right) - 3}}{2} + \dfrac{1}{{x + 1}}\\
= \dfrac{{3\left( {x + 1} \right)}}{2} + \dfrac{1}{{x + 1}} - \dfrac{3}{2}\\
Do:x > - 1\\
BDT:Co - si:\\
\dfrac{{3\left( {x + 1} \right)}}{2} + \dfrac{1}{{x + 1}} \ge 2\sqrt {\dfrac{{3\left( {x + 1} \right)}}{2}.\dfrac{1}{{x + 1}}} = \sqrt 6 \\
\to \dfrac{{3\left( {x + 1} \right)}}{2} + \dfrac{1}{{x + 1}} - \dfrac{3}{2} \ge \sqrt 6 - \dfrac{3}{2}\\
\to Min = \sqrt 6 - \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{2} = \dfrac{1}{{x + 1}}\\
\to {\left( {x + 1} \right)^2} = \dfrac{2}{3}\\
\to \left[ \begin{array}{l}
x = \sqrt {\dfrac{2}{3}} - 1\\
x = - \sqrt {\dfrac{2}{3}} - 1\left( l \right)
\end{array} \right.\\
b)y = \dfrac{{{x^3}}}{1} + \dfrac{2}{{{x^3}}}\\
Xét:x > 0\\
BDT:Co - si:\dfrac{{{x^3}}}{1} + \dfrac{2}{{{x^3}}} \ge 2\sqrt {\dfrac{{{x^3}}}{1}.\dfrac{2}{{{x^3}}}} = 2\sqrt 2 \\
\to Min = 2\sqrt 2 \\
\Leftrightarrow \dfrac{{{x^3}}}{1} = \dfrac{2}{{{x^3}}}\\
\to {x^6} = 2\\
\to x = \sqrt[6]{2}
\end{array}\)
