Đáp án:
$\text{(x,y,z)=(1,2,3)}$
Giải thích các bước giải:
$\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{1}{2}(x+y+z)$
$ĐKXĐ:\begin{cases}x \geq 0\\y-1 \geq 0\\z-2 \geq 0\\\end{cases}$
$↔\begin{cases}x \geq 0\\y \geq 1\\z \geq 2\\\end{cases}$
$pt↔2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=(x+y+z)$
$↔x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0$
$↔x-2\sqrt{x}+1+(y-1)-2\sqrt{y-1}+1+(z-2)-2\sqrt{z-2}+1=0$
$↔(\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$
$\text{vì} \begin{cases}(\sqrt{x}-1)^2 \geq 0\\(\sqrt{y-1}-1)^2 \geq 0\\(\sqrt{z-2}-1)^2 \geq 0\\\end{cases}$
$→(\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2 \geq 0$
$\text{mà đề bài cho} (\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$
$→\begin{cases}(\sqrt{x}-1)^2=0\\(\sqrt{y-1}-1)^2=0\\(\sqrt{z-2}-1)^2=0\\\end{cases}$
$↔\begin{cases}\sqrt{x}-1=0\\\sqrt{y-1}-1=0\\\sqrt{z-2}-1=0\\\end{cases}$
$↔\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\\\end{cases}$
$↔\begin{cases}x=1\\y-1=1\\z-2=1\\\end{cases}$
$↔\begin{cases}x=1(TMĐKXĐ)\\y=2(TMĐKXĐ)\\z=3(TMĐKXĐ)\\\end{cases}$
$\text{vậy (x,y,z)=(1,2,3)}$
