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Ta thấy : \(\left(2x-1\right)^{2008}\ge0\)
\(\left(y-\frac{2}{5}\right)^{2008}\ge0\)
\(\left|x+y+z\right|\ge0\)
Để \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y=z=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=-x-y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{-1}{2}-\frac{2}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)
\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)
Vậy.....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)
Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)
Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)
\(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)
Vậy...........
![](https://rs.olm.vn/images/avt/0.png?1311)