Question

Tìm x:

\(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\)

Answer 1

`( 3x - 1 ) . ( [-1] / 2 x + 5 ) = 0`

`@TH1: 3x - 1 = 0 => 3x = 1 => x = 1 : 3 = 1 / 3`

`@TH2: [-1] / 2 x + 5 = 0 => [-1] / 2 x = -5 => x = -5 : [-1] / 2 = 10`

Answer 2

\(\left(3x-1\right).\left(-\dfrac{1}{2}x+5\right)=0\)

\(\rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}3x=1\\-\dfrac{1}{2}x=-5\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

Answer 3

lớp 6 học cái nài :D?

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1\\\dfrac{-1}{2}x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)