(5x-1)(2x-1/3)=0
=>\(\left\{{}\begin{matrix}5x-1=0=>x=\dfrac{1}{5}\\2x-\dfrac{1}{3}=0=>x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy x=\(\dfrac{1}{5};\dfrac{1}{6}\)
\(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy ....
(5x-1).(2x-1/3)=0 <=> 5x-1=0 hoặc 2x-1/3 = 0
5x - 1 = 0 => 5x = 1(số bị trừ bằng hiệu + số trừ)
=> x = 1/5 ( thừa số bằng tích chia thừa số kia)
2x-1/3 = 0 tương tự trên x = 1/6
Ta có 2 trường hợp :
TH 1: TH2:
5x-1=0 2x - \(\dfrac{1}{3}\)=0
5x =0+1 2x = 0+ \(\dfrac{1}{3}\)
5x =1 2x = \(\dfrac{1}{3}\)
x = 1:5 x = \(\dfrac{1}{3}\):2
x = \(\dfrac{1}{5}\) x = \(\dfrac{1}{6}\)
