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\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
<=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|=0\)
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
Ta thấy : VT >= 0
Dấu "=" xảy ra <=> x-\(\sqrt{2}\)= 0 ; y+\(\sqrt{2}\)= 0 ; x+y+z = 0
<=> x=\(\sqrt{2}\); y=\(-\sqrt{2}\); z = 0
Vậy ...........
Tk mk nha
Ta có: \(\sqrt{\left(x-\sqrt{2}\right)^2}\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}\ge0;\left|x+y+z\right|\ge0\)
Mà theo đề: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
=> \(\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{\left(y+\sqrt{2}\right)^2}=\left|x+y+z\right|=0\)
=> \(x-\sqrt{2}=y+\sqrt{2}=x+y+z=0\)
=> \(x=\sqrt{2};y=-\sqrt{2};z=0\).
Vì \(\hept{\begin{cases}\sqrt{\left(x-\sqrt{2}\right)^2}\ge0\forall x\\\sqrt{\left(y+\sqrt{2}\right)^2}\ge0\forall y\\\left|x+y+z\right|\ge0\forall x;y;z\end{cases}}\)
Do đó : \(\hept{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{cases}}\)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(x-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-\sqrt{2}=0\\x+y+z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x+y=-z\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x=-z-y\end{cases}}\)
\(\sqrt{\left(x-3\sqrt{5}\right)^2}+\sqrt{\left(y+3\sqrt{5}\right)^2}+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left|x-3\sqrt{5}\right|+\left|y+3\sqrt{5}\right|+\left|x+y+z\right|=0\)
\(\Leftrightarrow\begin{cases}x-3\sqrt{5}=0\\y+3\sqrt{5}=0\\x+y+z=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=3\sqrt{5}\\y=-3\sqrt{5}\\z=-x-y=-3\sqrt{5}+3\sqrt{5}=0\end{cases}\)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|\)
Ta thấy: \(\begin{cases}\sqrt{\left(x-\sqrt{2}\right)^2}\ge0\\\sqrt{\left(y+\sqrt{2}\right)^2}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|\ge0\)
\(\Rightarrow\begin{cases}\sqrt{\left(x-\sqrt{2}\right)^2}=0\\\sqrt{\left(y+\sqrt{2}\right)^2}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}\left|x-\sqrt{2}\right|=0\\\left|y+\sqrt{2}\right|=0\\\left|x+y+z\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\\sqrt{2}+\left(-\sqrt{2}\right)+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{cases}\)