Question

\(\left(2x-1\right)\cdot\left(15-x\right)=0\)

Answer 1

`( 2x - 1 ) . ( 15 - x ) = 0`

`@TH1: 2x - 1 = 0`

        `=> 2x = 1`

        `=> x = 1 / 2`

`@TH2: 15 - x = 0`

          `=> x = 15`

Vậy `x = 1 / 2` hoặc `x = 15`

Answer 2

\(\left(2x-1\right)\cdot\left(15-x\right)=0\)

\(TH1:2x-1=0\)

           \(2x=0+1\)

           \(2x=1\)

            \(x=1:2\)

            \(x=\dfrac{1}{2}\)

\(TH2:15-x=0\)

                   \(x=15-0\)

                   \(x=15\)

Vậy \(x=\left[{}\begin{matrix}\dfrac{1}{2}\\15\end{matrix}\right.\)