\(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\-\dfrac{1}{2}x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
Vậy......................................
\(\left(3x-1\right).\left(\dfrac{-1}{2}.x+5\right)=0\)
Để biểu thức bằng 0 thì \(\left(3x-1\right)\) hoặc \(\left(\dfrac{-1}{2}.x+5\right)\) bằng 0
TH1:
\(3x-1=0\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
TH2:
\(\dfrac{-1}{2}.x+5=0\)
\(\dfrac{-1}{2}.x=-5\)
\(x=-5:\dfrac{-1}{2}\)
\(x=10\)
Vậy \(x=\left\{\dfrac{1}{3};10\right\}\)
Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=10\)
\(\left(3x+1\right)\cdot\left(\dfrac{-1}{2}\cdot x+5\right)=0\)
TH1: 3x+1=0
3x=0-1
3x=-1
x=-1:3
x=\(-\dfrac{1}{3}\)
TH2: \(\dfrac{-1}{2}\cdot x+5\)=0
\(\dfrac{-1}{2}\cdot x\)=0-5
\(\dfrac{-1}{2}\cdot x\)=-5
x=-5:\(\dfrac{-1}{2}\)
x=10
Vậy \(x\in\left\{\dfrac{-1}{3};10\right\}\).
