Giải:
Đặt:
\(\sqrt{x-2}=a\)
\(\sqrt{y-3}=b\)
\(\sqrt{z-5}=c\)
Đk: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\\z\ge5\end{matrix}\right.\)
Ta có:
\(A=\dfrac{x+y+z+4}{2}\)
\(\Leftrightarrow A=\dfrac{x-2+y-3+z-5+14}{2}\)
\(\Leftrightarrow A=\dfrac{a^2+b^2+c^2+14}{2}\)
\(\Leftrightarrow A=a+2b+3c\)
\(\Leftrightarrow A=\dfrac{a^2}{2}-a+\dfrac{1}{2}+\dfrac{b^2}{2}-2b+2+\dfrac{c^2}{2}-3c+\dfrac{9}{2}=0\)
\(\Leftrightarrow A=\left(\dfrac{a}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\right)^2+\left(\dfrac{b}{\sqrt{2}}-\sqrt{2}\right)^2+\left(\dfrac{c}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}}\right)=0\)
\(\Leftrightarrow a=1\Leftrightarrow x=3\Leftrightarrow y=b+3=7\Leftrightarrow z=c+5=14\)
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