\(x.5=x^2\)
\(\Rightarrow x^2-5x=0\)
\(\Rightarrow x.\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{0;5\right\}\)
\(\left(3x-1\right)^{2017}=\left(3x-1\right)^{2018}\)
\(\Rightarrow\left(3x-1\right)^{2018}-\left(3x-1\right)^{2017}=0\)
\(\Rightarrow\left(3x-1\right)^{2017}.\left[\left(3x-1\right)-1\right]=0\)
\(\Rightarrow\left(3x-1\right)^{2017}.\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{2017}=0\\\left(3x-2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{3};\dfrac{2}{3}\right\}\)
\(\left(x-1\right)^{x+2}=\left(x-1\right)^x\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^x\)
\(\Rightarrow\left(x-1\right)^x.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^x=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
