Ta có: \(\dfrac{1}{2}x+150\%x=2020\)
\(\Leftrightarrow2x=2020\)
hay x=1010
Vậy: x=1010
Ta có: \(\dfrac{1}{2}x+150\%x=2020\)
\(\Leftrightarrow\dfrac{1}{2}x+\dfrac{3}{2}x=2020\)
\(\Leftrightarrow\dfrac{4}{2}x=2020\)
\(\Leftrightarrow2x=2020\)
\(\Leftrightarrow x=1010\)
Vậy x=1010
\(\dfrac{1}{2}x+150\%x=2020\)
\(x.\left(\dfrac{1}{2}+\dfrac{3}{2}\right)=2020\)
\(x.2=2020\)
\(x=2020:2\)
\(x=1010\)