Ta có: \(\left(3x-7\right)^{2005}=\left(3x-7\right)^{2003}\)
\(\Leftrightarrow\left(3x-7\right)^{2005}-\left(3x-7\right)^{2003}=0\)
\(\Leftrightarrow\left(3x-7\right)^{2003}\left[\left(3x-7\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(3x-7\right)^{2003}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}\)
\(\left(3x-7\right)^{2005}=\left(3x-7\right)^{2003}\)
\(\Rightarrow\left(3x-7\right)^{2005}-\left(3x-7\right)^{2003}=0\)
\(\Leftrightarrow\left(3x-7\right)^{2003}[\left(3x-7\right)^2-1]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2003}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-7=0\\3x-7=1\end{cases}}\)hoặc \(3x-7=-1\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)hoặc \(x=2\)
Vậy ...............................
