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\(P=\sqrt{y}\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}=\left(6-\sqrt{x}-\sqrt{z}\right)\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}\)
\(P=-x+6\sqrt{x}-2z+12z=-\left(\sqrt{x}-3\right)^2-2\left(\sqrt{z}-3\right)^2+27\le27\)
\(P_{max}=27\) khi \(\left(x;y;z\right)=\left(9;0;9\right)\)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|\)
Ta thấy: \(\begin{cases}\sqrt{\left(x-\sqrt{2}\right)^2}\ge0\\\sqrt{\left(y+\sqrt{2}\right)^2}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|\ge0\)
\(\Rightarrow\begin{cases}\sqrt{\left(x-\sqrt{2}\right)^2}=0\\\sqrt{\left(y+\sqrt{2}\right)^2}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}\left|x-\sqrt{2}\right|=0\\\left|y+\sqrt{2}\right|=0\\\left|x+y+z\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\\sqrt{2}+\left(-\sqrt{2}\right)+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{cases}\)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
<=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|=0\)
Vì \(\left|x-\sqrt{2}\right|\ge0;\left|y+\sqrt{2}\right|\ge0;\left|x+y+z\right|\ge0\)
=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi \(\left|x-\sqrt{2}\right|=\left|y+\sqrt{2}\right|=\left|x+y+z\right|=0\)
\(\left|x-\sqrt{2}\right|=0\Leftrightarrow x-\sqrt{2}=0\Leftrightarrow x=\sqrt{2};\left|y+\sqrt{2}\right|=0\Leftrightarrow y+\sqrt{2}=0\Leftrightarrow y=-\sqrt{2}\)
\(\left|x+y+z\right|=0\Leftrightarrow x+y+z=0\Leftrightarrow\sqrt{2}+\left(-\sqrt{2}\right)+z=0\Leftrightarrow z=0\)
Vậy .......
do căn >= 0 lx+y+zl >=0 nên vế trái >=0
mà vế trái =0 => từng cái =0
