Giải thích các bước giải:

Cách 1:

$\begin{array}{l}
a)A = x + \dfrac{1}{{4x}}\left( {DK:x > 0} \right)\\
 \ge 2\sqrt {x.\dfrac{1}{{4x}}} \left( {BDT:Cauchy} \right)\\
 = 2\sqrt {\dfrac{1}{4}} \\
 = 1\\
 \Rightarrow MinA = 1
\end{array}$

Dấu bằng xảy ra:

$\begin{array}{l}
 \Leftrightarrow x = \dfrac{1}{{4x}}\\
 \Leftrightarrow {x^2} = \dfrac{1}{4}\\
 \Leftrightarrow x = \dfrac{1}{2}\left( {Do:x > 0} \right)
\end{array}$

Vậy $MinA = 1 \Leftrightarrow x = \dfrac{1}{2}$

$\begin{array}{l}
b)B = x + \dfrac{1}{{x - 2}}\left( {DK:x > 2} \right)\\
 = \left( {x - 2 + \dfrac{1}{{x - 2}}} \right) + 2\\
 \ge 2\sqrt {\left( {x - 2} \right).\dfrac{1}{{x - 2}}}  + 2\\
 = 4
\end{array}$

$ \Rightarrow MinB = 4$

Dấu bằng xảy ra

$\begin{array}{l}
 \Leftrightarrow x - 2 = \dfrac{1}{{x - 2}}\\
 \Leftrightarrow {\left( {x - 2} \right)^2} = 1\\
 \Leftrightarrow x - 2 = 1\left( {Do:x - 2 > 0} \right)\\
 \Leftrightarrow x = 3
\end{array}$

Vậy $MinB = 4 \Leftrightarrow x = 3$

Cách 2:

a) Ta có:

$\begin{array}{l}
A - 1 = x + \dfrac{1}{{4x}} - 1\\
 = \dfrac{{4{x^2} - 4x + 1}}{{4x}}\\
 = \dfrac{{{{\left( {2x - 1} \right)}^2}}}{{4x}}
\end{array}$

$\begin{array}{l}
Do:{\left( {2x - 1} \right)^2} \ge 0,\forall x > 0\\
 \Rightarrow \dfrac{{{{\left( {2x - 1} \right)}^2}}}{{4x}} \ge 0,\forall x > 0\\
 \Rightarrow A - 1 \ge 0\\
 \Rightarrow A \ge 1
\end{array}$

Dấu bằng xảy ra ${ \Leftrightarrow {{\left( {2x - 1} \right)}^2} = 0 \Leftrightarrow x = \dfrac{1}{2}}$

Vậy $MinA = 1 \Leftrightarrow x = \dfrac{1}{2}$

b) Ta có:

$\begin{array}{l}
B = x + \dfrac{1}{{x - 2}}\left( {DK:x > 2} \right)\\
B - 4 = x + \dfrac{1}{{x - 2}} - 4\\
 = \dfrac{{\left( {x - 4} \right)\left( {x - 2} \right) + 1}}{{x - 2}}\\
 = \dfrac{{{x^2} - 6x + 9}}{{x - 2}}\\
 = \dfrac{{{{\left( {x - 3} \right)}^2}}}{{x - 2}}\\
Do:{\left( {x - 3} \right)^2} \ge 0,\forall x > 2\\
 \Rightarrow \dfrac{{{{\left( {x - 3} \right)}^2}}}{{x - 2}} \ge 0,\forall x > 2\\
 \Rightarrow B - 4 \ge 0\\
 \Rightarrow B \ge 4\\
 \Rightarrow MinB = 4
\end{array}$

Dấu bằng xảy ra $ \Leftrightarrow {\left( {x - 3} \right)^2} = 0 \Leftrightarrow x = 3$

Vậy $MinB = 4 \Leftrightarrow x = 3$