ĐK: \(x\ge0;y\ge3;z\ge2\)
\(\Leftrightarrow x+y+z-2\sqrt{x}-4\sqrt{y-3}-2\sqrt{z-2}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-2}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-3}=2\\\sqrt{z-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=7\\z=3\end{matrix}\right.\) (TMĐK)
Vậy phương trình có nghiệm duy nhất \(\left(x;y;z\right)=\left(1;7;3\right)\).
Đk: `x >=0, y >=3, z >=2`.
`2 sqrt x + 4 sqrt(y-3) + 2 sqrt(z-2) = x + y + z +1`.
`(x + 2 sqrt x) + (y + 4 sqrt(y-3)) + (z - 2 + 2 sqrt(z-2) + 1) = 0`
`(x + 2 sqrt x) + (y + 4 sqrt(y-3)) + (sqrt(z-2)+1)^2 = 0`
Vì `x >=0, y >=3, z >=2`.
`-> x + 2 sqrt x >=0`
`y + 4 sqrt(y-3) >= y >=3`.
`(sqrt(z-2)+1)^2 >=0`.
`-> A >= 0 + 3 + 0 > 0`
Vậy không có `x, y, z` thỏa mãn
