a, xem lại đề , sửa rồi thì báo cho tui
b, \(\left(x+\frac{2}{5}\right)^5=\left(x+\frac{2}{5}\right)^3\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^5-\left(x+\frac{2}{5}\right)^3=0\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^3.\left[\left(x+\frac{2}{5}\right)^2-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+\frac{2}{5}\right)^3=0\\\left(x+\frac{2}{5}\right)^2-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{2}{5}\\\left(x+\frac{2}{5}\right)^2=1\end{cases}}}\)
Ta có \(\left(x+\frac{2}{5}\right)^2=1\)
\(\Rightarrow\hept{\begin{cases}x+\frac{2}{5}=1\\x+\frac{2}{5}=-1\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\x=-\frac{7}{5}\end{cases}}}\)
Vậy \(x\in\text{{}-\frac{2}{5};\frac{3}{5};-\frac{7}{5} \)}