Đáp án:

$x=3\sqrt5,y=-3\sqrt5,z=0$.

Giải thích các bước giải:

$\sqrt{\left.(x-3\sqrt5)\right.^{\!2}}+\sqrt{(\left.y+3\sqrt5)\right.^{\!2}}+|x+y+z|=0\\\Rightarrow|x-3\sqrt5|+|y+3\sqrt5|+|x+y+z|=0$

Ta có: $\left\{\begin{array}{l}|x-3\sqrt5|\ge0\\|y+3\sqrt5|\ge0\\|x+y+z|\ge0\end{array}\right.\forall\,x,y,z\in\mathbb R$

$\Rightarrow|x-3\sqrt5|+|y+3\sqrt5|+|x+y+z|\ge0\forall\,x,y,z\in\mathbb R$

Mà $|x-3\sqrt5|+|y+3\sqrt5|+|x+y+z|=0$

$\Leftrightarrow\left\{\begin{array}{l}|x-3\sqrt5|=0\\|y+3\sqrt5|=0\\|x+y+z|=0\end{array}\right.\\\Rightarrow\left\{\begin{array}{l}x-3\sqrt5=0\\y+3\sqrt5=0\\x+y+z=0\end{array}\right.\\\Rightarrow\left\{\begin{array}{l}x=3\sqrt5\\y=-3\sqrt5\\3\sqrt5+(-3\sqrt5)+z=0\end{array}\right.\\\Rightarrow\left\{\begin{array}{l}x=3\sqrt5\\y=-3\sqrt5\\z=0\end{array}\right.$

Vậy $x=3\sqrt5,y=-3\sqrt5,z=0$.

Lời giải chi tiết:

`\sqrt{(x-3\sqrt{5})^2}+\sqrt{(y+3\sqrt{5})^2}+|x+y+z|=0`

Ta có`:` `{(\sqrt{(x-3\sqrt{5})^2}\ge0∀x),(\sqrt{(y+3\sqrt{5})^2}\ge0∀y),(|x+y+z|\ge0∀x\text{,}y\text{,}z):}`

`=>``\sqrt{(x-3\sqrt{5})^2}+\sqrt{(y+3\sqrt{5})^2}+|x+y+z|\ge0∀x,y,z`

Dấu "`=`" xảy ra khi`:`

`{(\sqrt{(x-3\sqrt{5})^2}=0),(\sqrt{(y+3\sqrt{5})^2}=0),(|x+y+z|=0):}`

`=>{(x-3\sqrt{5}=0),(y+3\sqrt{5}=0),(x+y+z=0):}`

`=>{(x=3\sqrt{5}),(y=-3\sqrt{5}),(3\sqrt{5}+(-3\sqrt{5})+z=0):}`

`=>{(x=3\sqrt{5}),(y=-3\sqrt{5}),(z=0):}`

Vậy `(x;y;z)=(3\sqrt{5};-3\sqrt{5};0)`