Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \sqrt {x + 2\sqrt {2x - 4} } + \sqrt {x - 2\sqrt {2x - 4} } \\
= \sqrt {\dfrac{1}{2}.\left( {2x + 4\sqrt {2x - 4} } \right)} + \sqrt {\dfrac{1}{2}\left( {2x - 4\sqrt {2x - 4} } \right)} \\
= \sqrt {\dfrac{1}{2}.\left[ {\left( {2x - 4} \right) + 4\sqrt {2x - 4} + 4} \right]} + \sqrt {\dfrac{1}{2}.\left[ {\left( {2x - 4} \right) - 4\sqrt {2x - 4} + 4} \right]} \\
= \sqrt {\dfrac{1}{2}.{{\left( {\sqrt {2x - 4} + 2} \right)}^2}} + \sqrt {\dfrac{1}{2}.{{\left( {\sqrt {2x - 4} - 2} \right)}^2}} \\
= \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt {2x - 4} + 2} \right| + \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt {2x - 4} - 2} \right|\\
= \dfrac{1}{{\sqrt 2 }}.\left[ {\sqrt {2x - 4} + 2 + \left| {\sqrt {2x - 4} - 2} \right|} \right]\\
TH1:\,\,\,\,\sqrt {2x - 4} - 2 \le 0 \Leftrightarrow 0 \le 2x - 4 \le 4 \Leftrightarrow 2 \le x \le 4\\
\Rightarrow P = \dfrac{1}{{\sqrt 2 }}.\left[ {\sqrt {2x - 4} + 2 + \left( {2 - \sqrt {2x - 4} } \right)} \right] = \dfrac{1}{{\sqrt 2 }}.4 = 2\sqrt 2 \\
TH2:\,\,\,\,\sqrt {2x - 4} - 2 > 0 \Leftrightarrow x > 4\\
\Rightarrow P = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {2x - 4} + 2 + \sqrt {2x - 4} - 2} \right) = \dfrac{1}{{\sqrt 2 }}.2\sqrt {2x - 4} = \sqrt 2 .\sqrt {2x - 4} \\
\sqrt {2x - 4} > 2 \Rightarrow P > \sqrt 2 .2 = 2\sqrt 2 \\
\Rightarrow {P_{\min }} = 2\sqrt 2 \Leftrightarrow 2 \le x \le 4
\end{array}\)
