Giải thích các bước giải:
$\begin{array}{l}
a)y = 2x + \dfrac{3}{{x + 2}}\left( {DK:x > - 2} \right)\\
= 2\left( {x + 2} \right) + \dfrac{3}{{x + 2}} - 4\\
= \left( {2\left( {x + 2} \right) + \dfrac{3}{{x + 2}}} \right) - 4\\
\ge 2\sqrt {2\left( {x + 2} \right).\dfrac{3}{{x + 2}}} - 4\left( {B{\rm{D}}T:Cauchy} \right)\\
= 2\sqrt 6 - 4\\
\Rightarrow Miny = 2\sqrt 6 - 4
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow 2\left( {x + 2} \right) = \dfrac{3}{{x + 2}}\\
\Leftrightarrow x + 2 = \dfrac{{\sqrt 6 }}{2}\left( {Do:x > - 2} \right)\\
\Leftrightarrow x = \dfrac{{\sqrt 6 }}{2} - 2
\end{array}$
Vậy $Miny = 2\sqrt 6 - 4 \Leftrightarrow x = \dfrac{{\sqrt 6 }}{2} - 2$
$\begin{array}{l}
b)y = \dfrac{x}{{1 - x}} + \dfrac{5}{x}\left( {DK:0 < x < 1} \right)\\
= \dfrac{x}{{1 - x}} + \dfrac{5}{x} - 5 + 5\\
= \dfrac{x}{{1 - x}} + \dfrac{{5\left( {1 - x} \right)}}{x} + 5\\
\ge 2\sqrt {\dfrac{x}{{1 - x}}.\dfrac{{5\left( {1 - x} \right)}}{x}} + 5\left( {BDT:Cauchy} \right)\\
= 2\sqrt 5 + 5
\end{array}$
$ \Rightarrow Miny = 2\sqrt 5 + 5$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \dfrac{x}{{1 - x}} = \dfrac{{5\left( {1 - x} \right)}}{x}\\
\Leftrightarrow {x^2} = 5{\left( {1 - x} \right)^2}\\
\Leftrightarrow x = \sqrt 5 \left( {1 - x} \right)\\
\Leftrightarrow x = \dfrac{{\sqrt 5 }}{{1 + \sqrt 5 }}
\end{array}$
Vậy $Miny = 2\sqrt 5 + 5 \Leftrightarrow x = \dfrac{{\sqrt 5 }}{{1 + \sqrt 5 }}$
