a,ĐKXĐ: \(\forall x\in R\)
\(2x-18x^3=0\\ \Leftrightarrow-\left(18x^3-2x\right)=0\\ \Leftrightarrow18x^3-2x=0\\ \Leftrightarrow2x\left(9x^2-1\right)=0\\ \Leftrightarrow2x\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{3}\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\frac{1}{3};-\frac{1}{3}\right\}\)
b, \(5x\left(x-2\right)=x-2\)
ĐKXĐ: \(\forall x\in R\)
\(\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\frac{1}{5}\right\}\)
