Đặt :
\(A=\left(5x-1\right)\cdot\left(2x-\dfrac{1}{3}\right)=0\)
Để A = 0 <=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy để GTBT trên = 0 \(\Leftrightarrow x=\dfrac{1}{5}\) hoặc \(x=\dfrac{1}{6}\)
\(\left(5\times x-1\right)\left(2\times x-\dfrac{1}{3}\right)=0\Leftrightarrow5\times x-1=0\) hoặc \(2\times x-\dfrac{1}{3}=0\)
Nếu 5.x-1=0
5x=0+1
5x=1
=>x=1:5
=>x=\(\dfrac{1}{5}\)
Nếu \(2.x-\dfrac{1}{3}=0\)
2.x=\(0+\dfrac{1}{3}\)
2.x=\(\dfrac{1}{3}\)
=>x=\(\dfrac{1}{3}:2\)
=>\(x=\dfrac{1}{6}\)
Vậy \(x=\dfrac{1}{5}\) hoặc \(x=\dfrac{1}{6}\)
