Gọi st1: \(x\) ; \(st2:y;st3:z\)
Ta có: \(x+y+z=240\)
\(\dfrac{2}{3}x=\dfrac{1}{2}y=\dfrac{2}{5}z\)
\(\dfrac{2}{3}x=\dfrac{1}{2}y=>y=\dfrac{2}{3}x:\dfrac{1}{2}=\dfrac{4}{3}x\)
\(\dfrac{2}{3}x=\dfrac{2}{5}z=>z=\dfrac{2}{3}x:\dfrac{2}{5}=\dfrac{5}{3}x\)
\(=>x+\dfrac{4}{3}x+\dfrac{5}{3}x=240\)
\(=>4x=240\)
\(=>x=60\)
\(=>y=60\times\dfrac{4}{3}=80\)
\(=>z=60\times\dfrac{5}{3}=100\)
Vậy st1 là 60 ; st2 là 80 ; st3 là 100
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