Question

[Lê Thúy Hằng]

Tìm x, y, z biết:

\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\)I x + y + z I         =0

Answer 1

[Bùi Hữu Vinh]

\(\left(2x-1\right)^{2008}\ge0với\forall x\) mà,\(\left(y-\frac{2}{5}\right)^{2008}\ge0với\forall y\)lại có\(|x+y+z|\ge0với\forall x,y,z\)

\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0với\forall x,y,z\)Dấu ''='' xảy ra khi \(\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=-\frac{9}{10}\end{cases}}}\)

Answer 2

[Yêu nè]

\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)+\left|x+y+z\right|=0\)

Ta có \(\hept{\begin{cases}\left(2x-1\right)^{2008}\ge0\forall x\\y-\frac{2}{5}\ge0\forall y\\\left|x+y+z\right|\ge0\forall x;y;z\end{cases}}\)

 => \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)+\left|x+y+z\right|\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(2x-1\right)^{2008}=0\\y-\frac{2}{5}=0\\\left|x+y+z\right|=0\end{cases}}\)

                         \(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y=\frac{2}{5}\\x+y+z=0\end{cases}}\)

                        \(\Leftrightarrow\hept{\begin{cases}2x=1\\y=\frac{2}{5}\\x+y+z=0\end{cases}}\)

                          \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}}\)

                         \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=0-\frac{1}{2}-\frac{2}{5}=\frac{-5}{10}-\frac{4}{10}=\frac{-9}{10}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{2}{5};z=\frac{-9}{10}\)

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