\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\left[\begin{matrix}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}3x-1=0\\3x-1=\pm1\end{matrix}\right.\)
+) \(3x-1=0\Rightarrow x=\frac{1}{3}\)
+) \(\left[\begin{matrix}3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{2}{3}\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}\)
(3x−1)10=(3x−1)20
=>(3x−1)10=(3x−1)10.(3x−1)10
(3x−1)10:(3x−1)10=(3x−1)10
vậy (3x−1)10=\(1^{10}\)
ta có (3x−1)10=1 hoặc (3x−1)10=-1
=>TH1 3x-1=1 TH2 :3x-1=-1
3x=1+1 3x=-1+1
3x=2 3x=0
x=\(\frac{2}{3}\) x=0
vậy x=\(\frac{2}{3}\) hoặc x=0
