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\(\left(x-2008\right)^{2016}+\left(y-2009\right)^{2017}=0\)

\(\Rightarrow x-2008=0\Rightarrow x=2008\)

\(\Rightarrow y-2009=0\Rightarrow y=2009\)

\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)

\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)

\(\frac{x}{2008}=1\)

x=2008

ban tien lam dung roi do.ket qua x=2008.k cho mk nha

Sửa đề; \(\left(\frac{3x-5}{9}\right)^{2006}+\left(\frac{3y+0,4}{3}\right)^{2008}\le0\)

Ta có: \(\left(\frac{3x-5}{9}\right)^{2006}\ge0\forall x\)

\(\left(\frac{3y+0,4}{3}\right)^{2008}\ge0\forall y\)

Do đó: \(\left(\frac{3x-5}{9}\right)^{2006}+\left(\frac{3y+0,4}{3}\right)^{2008}\ge0\forall x,y\)

mà \(\left(\frac{3x-5}{9}\right)^{2006}+\left(\frac{3y+0,4}{3}\right)^{2008}\le0\)

nên dấu '=' xảy ra khi \(\begin{cases}3x-5=0\\ 3y+0,4=0\end{cases}\Rightarrow\begin{cases}3x=5\\ 3y=-0,4=-\frac25\end{cases}\Rightarrow\begin{cases}x=\frac53\\ y=-\frac{2}{15}\end{cases}\)

Ta có: \(\left(3x-5\right)^{2006}\ge0\forall x\)

\(\left(y^2-1\right)^{2008}\ge0\forall y\)

\(\left(x-z\right)^{2010}\ge0\forall x,z\)

Do đó: \(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2010}>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}3x-5=0\\ y^2-1=0\\ z-x=0\end{cases}\Rightarrow\begin{cases}x=\frac53\\ z=x=\frac53\\ y^2=1\end{cases}\Rightarrow\begin{cases}x=\frac53\\ z=\frac53\\ y\in\left\lbrace1;-1\right\rbrace\end{cases}\)