Đáp án:
$\sqrt P = 1$
Giải thích các bước giải:
ĐK: $x,y,z>0$
Ta có: $xyz=4$
Khi đó:
$\begin{array}{l}
P = \dfrac{{\sqrt x }}{{\sqrt {xy} + \sqrt x + 2}} + \dfrac{{\sqrt y }}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{2\sqrt z }}{{\sqrt {xz} + 2\sqrt z + 2}}\\
= \dfrac{{\sqrt x }}{{\sqrt {xy} + \sqrt x + \sqrt {xyz} }} + \dfrac{{\sqrt y }}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt z }}{{\dfrac{1}{2}\sqrt {xz} + \sqrt z + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt y + 1 + \sqrt {yz} } \right)}} + \dfrac{{\sqrt y }}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt z }}{{\dfrac{{\sqrt {xz} }}{{\sqrt {xyz} }} + \sqrt z + 1}}\\
= \dfrac{1}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt y }}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt z }}{{\dfrac{1}{{\sqrt y }} + \sqrt z + 1}}\\
= \dfrac{1}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt y }}{{\sqrt {yz} + \sqrt y + 1}} + \dfrac{{\sqrt {yz} }}{{1 + \sqrt {yz} + \sqrt y }}\\
= \dfrac{{1 + \sqrt y + \sqrt {yz} }}{{\sqrt {yz} + \sqrt y + 1}}\\
= 1\\
\Rightarrow P = 1\\
\Rightarrow \sqrt P = 1
\end{array}$
Vậy $\sqrt P = 1$
