\(\left(3x+1\right)^9=\left(3x+1\right)^2\)
\(\left(3x+1\right)^9-\left(3x+1\right)^2=0\)
\(\left(3x+1\right)^2\cdot\left(3x+1\right)^7-\left(3x+1\right)^2=0\)
\(\left(3x+1\right)^2\left[\left(3x+1\right)^7-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x+1\right)^2=0\\\left(3x+1\right)^7-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\\left(3x+1\right)^7=1=1^7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-1\\3x+1=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=0\end{cases}}\)
Vậy x = { -1/3; 0 }
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