\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
\(\Leftrightarrow\left(2x-3\right)^4-\left(2x-3\right)^6=0\)
\(\Leftrightarrow\left(2x-3\right)^4\cdot\left[1-\left(2x-3\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-3\right)^4=0\\1-\left(2x-3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-\left(2x-3\right)^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\\left(2x-3\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x-3=1\\2x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy \(x_1=1;x_2=\dfrac{3}{2};x_3=2\)
\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
\(\Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\)
\(\Leftrightarrow\left(2x-4\right)^4\left[\left(2x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-4\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\)
+) \(\left(2x-4\right)^4=0\Leftrightarrow x=2\)
+) \(\left(2x-3\right)^2-1=0\Leftrightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy...
Ta có :
\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
\(\Leftrightarrow\) Ta có các trường hợp :
TH1 : \(2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
TH2 : \(2x-3=1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\left(TM\right)\)
TH3 : \(2x-3=-1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy \(x\in\left\{\dfrac{3}{2};1;2\right\}\) là giá trị cần tìm
