Vì mọi hạng tử trong đa thức đều lớn hơn hoặc bằng 0 nên ta xét 3 trường hợp:
(+) \(\left(2x-10\right)^{2008}=0\) \(\Rightarrow\) \(2x-1=0\)
\(\Rightarrow x=\frac{1}{2}\)
(+) \(\left(y-\frac{2}{5}\right)^{2008}\) \(\Rightarrow y-\frac{2}{5}=0\)
\(\Rightarrow y=\frac{2}{5}\)
(+) \(\left|x+y+z\right|=0\) \(\Rightarrow x+y+z=0\)
\(\Rightarrow\) \(\frac{1}{2}+\frac{2}{5}+z=0\)
\(\Rightarrow\) \(\frac{7}{5}+z=0\)
\(\Rightarrow z=-\frac{7}{5}\)
Vì\(\hept{\begin{cases}\left(2x-1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y-z\right|\ge0\end{cases}}\)
=>\(\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\hept{\begin{cases}2x=1\\y=\frac{2}{5}\\x+y-z=0\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}\end{cases}}\)
KL: (x,y,z)=(\(\frac{1}{2};\frac{2}{5};\frac{9}{10}\))
Với mọi \(x,y,z\), ta có:
\(\Leftrightarrow\)\(\left(2x-1\right)^{2008}\ge0\)
\(\Leftrightarrow\)\(\left(y-\frac{2}{5}\right)^{2008}\ge0\)
\(\left|x+y-z\right|\ge0\)
Để: \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)thì:
\(2x-1=0\) \(y-\frac{2}{5}=0\)
\(\Rightarrow\)\(2x=1\) \(y=0+\frac{2}{5}\)
\(\Rightarrow\)\(x=\frac{1}{2}\) \(y=\frac{2}{5}\)
\(x+y-z=0\Rightarrow z=x+y=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}\)
Vậy: \(x=\frac{1}{2};y=\frac{2}{5};z=\frac{9}{10}\)
