Đáp án:
`(x,y,z)=(2010;-2007;3).`
Giải thích các bước giải:
`\sqrt{x-2009}+\sqrt{y+2008}+\sqrt{z-2}=1/2(x+y+z)`
Điều kiện:`{(x-2009>=0),(y+2008>=0),(z-2>=0):}`
`<=>{(x>=2009),(y>=-2008),(z>=2):}`
`PT<=>2\sqrt{x-2009}+2\sqrt{y+2008}+2\sqrt{z-2}=x+y+z`
`<=>x-2\sqrt{x-2009}+y-2\sqrt{y+2008}+z-2\sqrt{z-2}=0`
`<=>x-2009-2\sqrt{x-2009}+1+y+2008-2\sqrt{y+2008}+1+z-2-2\sqrt{z-2}+1=0`
`<=>(\sqrt{x-2009}-1)^2+(\sqrt{y+2008}-1)^2+(\sqrt{z-2}-1)^2=0`
Vì `{((\sqrt{x-2009}-1)^2>=0),((\sqrt{y+2008}-1)^2>=0),((\sqrt{z-2}-2)^1>=0):}`
`=>(\sqrt{x-2009}-1)^2+(\sqrt{y+2008}-1)^2+(\sqrt{z-2}-1)^2>=0`
Mà đề bài cho `(\sqrt{x-2009}-1)^2+(\sqrt{y+2008}-1)^2+(\sqrt{z-2}-1)^2=0`
`<=>{((\sqrt{x-2009}-1)^2=0),((\sqrt{y+2008}-1)^2=0),((\sqrt{z-2}-1)^2=0):}`
`<=>{(\sqrt{x-2009}-1=0),(\sqrt{y+2008}-1=0),(\sqrt{z-2}-1=0):}`
`<=>{(x-2009=1),(y+2008=1),(z-2=1):}`
`<=>{(x=2010(TM)),(y=-2007(TM)),(z=3(TM)):}`
Vậy `(x,y,z)=(2010;-2007;3).`
