Ta có:
\(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
= 5x(x-2000) - (x-2000) =0
(x-2000)(5x-1) =0
x -2000 = 0=> x=2000
5x-1 =0 => x = 1/5
Ta có: 5x(x-2000)-x+2000=0
5x(x-2000)+(x-2000)=0
(x-2000). (5x-1)=0
TH1: x-2000=0
x=2000
TH2: 5x-1= 0
x=1/5
Vậy x=2000 hoặc x=1/5
