\(\dfrac{3}{4}+\dfrac{1}{6}\times3+\dfrac{2}{5}:\dfrac{4}{15}\)
\(=\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{2}{5}\times\dfrac{15}{4}\)
\(=\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{3}{2}\) \(=\dfrac{3+2+3}{4}=\dfrac{8}{4}=2\)
Gọi số đó là \(\overline{abcd}\) \(\left(a\ne0\right)\)
Ta có: \(\overline{abcd}+a+b+c+d=2020\)
\(\Leftrightarrow1000a+100b+10c+d+a+b+c+d=2020\)
\(\Leftrightarrow1001a+101b+11c+2d=2020\)