xin lỗi mk ấn nhầm
<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
<=>\(\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
Vậy ...............
ĐKXĐ : \(x\ge1;y\ge2;z\ge3\)
Ta có [ (x - 1) - 2\(\sqrt{x-1}\) +1 ]+[ (y - 2) - 4\(\sqrt{y-2}\) +4]+[ (z - 3) - 6\(\sqrt{z-3}\) +9] = 0
\(\text{Điều kiện}:x \ge 1; y \ge 2; z \ge 3\)
\( x + y + z + 8 = 2\sqrt {x - 1} + 4\sqrt {y - 2} + 6\sqrt {z - 3} \\ \Leftrightarrow x - 2\sqrt {x - 1} + y - 4\sqrt {y - 2} + x - 6\sqrt {z - 3} + 8 = 0\\ \Leftrightarrow \left( {x - 1 - 2\sqrt {x - 1} + 1} \right) + \left( {y - 2 - 4\sqrt {y - 2} + 4} \right) + \left( {z - 3 - 6\sqrt {z - 3} + 9} \right) = 0\\ \Leftrightarrow {\left( {\sqrt {x - 1} - 1} \right)^2} + {\left( {\sqrt {y - 2} - 2} \right)^2} + {\left( {\sqrt {z - 3} - 3} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 1} - 1 = 0\\ \sqrt {y - 2} - 2 = 0\\ \sqrt {z - 3} - 3 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 6\\ z = 12 \end{array} \right. \)
