\(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}\)
\(\Rightarrow\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(\Rightarrow\left(3x-7\right)^{2017}\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\left(3x-7\right)^{2017}=0\text{ hoặc }\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow3x-7=0\text{ hoặc }\left(3x-7\right)^2=1\)
\(\Rightarrow3x-7=0\text{ hoặc } \hept{\begin{cases}3x-7=1\\3x-7=-1\end{cases}}\)
\(\Rightarrow3x=7\text{ hoặc }3x=8\text{ hoặc }3x=6\)
\(\Rightarrow x=\frac{7}{3}\text{ hoặc }x=\frac{8}{3}\text{ hoặc }x=2\)
\(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}\)
\(\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(\left(3x-7\right)^{2017}\cdot\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2017}=0\\\left(3x-7\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=7\\3x-7=\pm1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=6\text{ hoặc }3x=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=2\text{ hoặc }x=\frac{8}{3}\end{cases}}\)
\(\Rightarrow x\in\left\{\frac{7}{3};2;\frac{8}{3}\right\}\)
Bài giải
\(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}\)
\(\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(\left(3x-7\right)^{2017}\cdot\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2017}=0\\\left(3x-7\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=7\\3x-7=-1\text{ hoặc }3x-7=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=6\text{ hoặc }3x=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=2\text{ hoặc }x=\frac{8}{3}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{7}{3}\text{ ; }2\text{ ; }\frac{8}{3}\right\}\)
Nếu có điều kiện gì thì bạn chọn x nếu không thì chọn cả 3 đáp án như trên nha !
Ta có : \(\left(3x-7\right)^{2019}=\left(3x-7\right)^{2017}=>\left(3x-7\right)^{2019}-\left(3x-7\right)^{2017}=0\)
\(=>\left(3x-7\right)^{2017}\left[\left(3x-7\right)^2-1\right]=0\)
\(=>\orbr{\begin{cases}(3x-7)^{2017}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)
\(=>\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3} hoac x=2\end{cases}}\)\(=>x=\frac{7}{3}\)hoặc\(x=\frac{8}{3}\)hoặc\(x=2\)
Vậy ...
\(=>\orbr{\begin{cases}3x=7\\\orbr{\begin{cases}3x-7=1\\3x-7=-1\end{cases}}\end{cases}}\)
\(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}=>\orbr{\begin{cases}3x-7=0\\\orbr{\begin{cases}3x-7=1\\3x-7=-1\end{cases}}\end{cases}}}\)
\(=>\orbr{\begin{cases}\left(3x-7\right)^{2017}=0\\\left(3x-7\right)^2-1=0\end{cases}=>\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}=>\orbr{\begin{cases}3x=7\\\orbr{\begin{cases}3x-7=1\\3x-7=-1\end{cases}}\end{cases}=>\orbr{\begin{cases}x=\frac{7}{3}\\\orbr{\begin{cases}\frac{8}{3}\\x=2\end{cases}}\end{cases}}}}}\)