Giải thích các bước giải:

Áp dụng bất đẳng thức Cô si ta có:

\(\begin{array}{l}
a,\\
A = \dfrac{x}{2} + \dfrac{{18}}{x} \ge 2\sqrt {\dfrac{x}{2}.\dfrac{{18}}{x}}  = 2\sqrt 9  = 6\\
{A_{\min }} = 6 \Leftrightarrow \dfrac{x}{2} = \dfrac{{18}}{x} \Leftrightarrow x = 6\\
b,\\
B = \dfrac{x}{2} + \dfrac{2}{{x - 1}} = \left( {\dfrac{x}{2} - \dfrac{1}{2}} \right) + \dfrac{2}{{x - 1}} + \dfrac{1}{2}\\
 = \dfrac{{x - 1}}{2} + \dfrac{2}{{x - 1}} + \dfrac{1}{2} \ge 2\sqrt {\dfrac{{x - 1}}{2}.\dfrac{2}{{x - 1}}}  + \dfrac{1}{2} = 2.\sqrt 1  + \dfrac{1}{2} = \dfrac{5}{2}\\
 \Rightarrow {B_{\min }} = \dfrac{5}{2} \Leftrightarrow \dfrac{{x - 1}}{2} = \dfrac{2}{{x - 1}} \Leftrightarrow x - 1 = 2 \Leftrightarrow x = 3\\
c,\\
C = \dfrac{{3x}}{2} + \dfrac{1}{{x + 1}} = \left( {\dfrac{{3x}}{2} + \dfrac{3}{2}} \right) + \dfrac{1}{{x + 1}} - \dfrac{3}{2}\\
 = \dfrac{{3.\left( {x + 1} \right)}}{2} + \dfrac{1}{{x + 1}} - \dfrac{3}{2} \ge 2.\sqrt {\dfrac{{3\left( {x + 1} \right)}}{2}.\dfrac{1}{{x + 1}}}  - \dfrac{3}{2} = 2\sqrt {\dfrac{3}{2}}  - \dfrac{3}{2}\\
{C_{\min }} = 2\sqrt {\dfrac{3}{2}}  - \dfrac{3}{2} \Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{2} = \dfrac{1}{{x + 1}} \Leftrightarrow x + 1 = \sqrt {\dfrac{2}{3}}  \Leftrightarrow x = \sqrt {\dfrac{2}{3}}  - 1\\
d,\\
D = \dfrac{x}{3} + \dfrac{5}{{2x - 1}} = \dfrac{{2x}}{6} + \dfrac{5}{{2x - 1}} = \left( {\dfrac{{2x}}{6} - \dfrac{1}{6}} \right) + \dfrac{5}{{2x - 1}} + \dfrac{1}{6}\\
 = \dfrac{{2x - 1}}{6} + \dfrac{5}{{2x - 1}} + \dfrac{1}{6} \ge 2\sqrt {\dfrac{{2x - 1}}{6}.\dfrac{5}{{2x - 1}}}  + \dfrac{1}{6} = 2\sqrt {\dfrac{5}{6}}  + \dfrac{1}{6}\\
 \Rightarrow {D_{\min }} = 2\sqrt {\dfrac{5}{6}}  + \dfrac{1}{6} \Leftrightarrow \dfrac{{2x - 1}}{6} = \dfrac{5}{{2x - 1}} \Leftrightarrow 2x - 1 = \sqrt {30}  \Leftrightarrow x = \dfrac{{\sqrt {3x}  + 1}}{2}\\
e,\\
E = \dfrac{x}{{1 - x}} + \dfrac{5}{x} = \dfrac{x}{{1 - x}} + \left( {\dfrac{5}{x} - 5} \right) + 5\\
 = \dfrac{x}{{1 - x}} + \dfrac{{5 - 5x}}{x} + 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 = 2\sqrt 5  + 5\\
 \Rightarrow {E_{\min }} = 2\sqrt 5  + 5 \Leftrightarrow \dfrac{x}{{1 - x}} = \dfrac{{5\left( {1 - x} \right)}}{x}
\end{array}\)