(3x-7)2007 = (3x-7)2005
=> (3x-7)2007 - (3x-7)2005 = 0
=> (3x-7)2005[(3x-7)2-1] = 0
\(\Rightarrow\left[{}\begin{matrix}\left(3x-7\right)^{2005}=0\\\left(3x-7\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-7=0\\\left(3x-7\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=7\\3x-7=\pm1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{8}{3}\\\dfrac{6}{3}\end{matrix}\right.\)Vậy:..........
\(\left(3x-7\right)^{2007}=\left(3x-7\right)^{2005}\)
Để \(\left(3x-7\right)^{2007}=\left(3x-7\right)^{2005}\)
thì 3x- 7= 1 hoặc 3x-7= 0
\(\left\{{}\begin{matrix}\left(3x-7\right)^{2007}=1\\\left(3x-7\right)^{2007}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-7=1\\3x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=8\\3x=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(3x-7\right)^{2005}=1\\\left(3x-7\right)^{2005}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-7=1\\3x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=8\\3x=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(\left(3x-7\right)^{2007}=\left(3x-7\right)^{2005}\\ \Rightarrow\left(3x-7\right)^{2007}-\left(3x-7\right)^{2005}=0\\ \Leftrightarrow\left(3x-7\right)^{2005}\left(\left(3x-7\right)^2-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-7=0\\\left(3x-7\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{8}{3}\\x=2\end{matrix}\right.\)
