Đáp án:

`ĐKXĐ : y ≥ 2000 ; z ≥ 2001 ; x ≥ 2002`

Ta có

`\sqrt{y - 2000} + \sqrt{z - 2001} +  \sqrt{x - 2002} = 1/2 (x + y + z) - 3000`

`<=> 2\sqrt{y - 2000} + 2\sqrt{z - 2001} + 2\sqrt{x - 2002} = x + y + z - 6000`

`<=> x + y + z - 2\sqrt{y - 2000} -  2\sqrt{z - 2001} -  2\sqrt{x - 2002} - 6000 = 0`

`<=> [(y - 2000) - 2\sqrt{y - 2000} + 1] + [(z - 2001) - 2\sqrt{z - 2001} + 1] + [(x - 2002) - 2\sqrt{x - 2002} + 1] = 0`

`<=> (\sqrt{y - 2000} - 1)^2 + (\sqrt{z - 2001} - 1)^2 + (\sqrt{x - 2002} - 1)^2 = 0`

`<=> {\sqrt{y - 2000}  - 1 = 0`

        `{\sqrt{z - 2001}  - 1 = 0`

        `{\sqrt{x - 2002} - 1 = 0`

`<=> {\sqrt{y - 2000}  = 1`

        `{\sqrt{z - 2001} = 1`

        `{\sqrt{x - 2002} = 1`

`<=> {y - 2000 = 1`

        `{z - 2001 = 1`

        `{x - 2002 = 1`

`<=> {y = 2001`

        `{z = 2002`

        `{x = 2003` 

Giải thích các bước giải:

`sqrt(y-2000)+sqrt(z-2001)+sqrt(x-2002)=1/2(x+y+z)-3000`

`<=>2sqrt(y-2000)+2sqrt(z-2001)+2sqrt(x-2002)=x+y+z-6000`

`<=>x+y+z-2sqrt(y-2000)-2sqrt(z-2001)-2sqrt(x-2002)-6000=0`

`<=>[(x-2002)-2sqrt(x-2002)+1]+[(y-2000)-2sqrt(y-2000)+1]+[(z-2001)-2sqrt(z-2001)+1]=0`

`<=>(sqrt(x-2002)-1)^2+(sqrt(y-2000)-1)^2+(sqrt(z-2001)-1)^2=0`

`<=>`$\begin{cases}\sqrt{x-2002}-1=0\\\sqrt{y-2000}-1=0\\\sqrt{z-2001}-1=0\end{cases}$

`<=>`$\begin{cases}\sqrt{x-2002}=1\\\sqrt{y-2000}=1\\\sqrt{z-2001}=1\end{cases}$

`<=>`$\begin{cases}x-2002=1\\y-2000=1\\z-2001=1\end{cases}$

`<=>`$\begin{cases}x=2003\\y=2001\\z=2002\end{cases}$