\(5x\left(x-1\right)=1-x\\ \Leftrightarrow5x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy...
\(5x\left(x-1\right)=1-x\\ \Leftrightarrow5x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\5x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
5x(x-1)-(1-x)=0
=> 5x(x-1) -(-(x-1))=0
=> (5x+1)(x-1)=0
=> 5x+1=0 hoặc x-1=0
=> x=(-1)/5 hoặc x=1
