Khi xuôi dòng: \(t=\dfrac{s}{v_{xuong\cdot bo}}=\dfrac{s}{v_{xuong\cdot nuoc}+4}\)
Khi ngược dòng: \(t'=\dfrac{s}{v_{xuong\cdot bo}}=\dfrac{s}{v_{xuong\cdot nuoc}-4}\)
Ta có hpt: \(\left\{{}\begin{matrix}4=\dfrac{s}{v_{xuong\cdot nuoc}+4}\\5=\dfrac{s}{v_{xuong\cdot nuoc}-4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}s=160km\\v_{xuong\cdot nuoc}=36\left(\dfrac{km}{h}\right)\end{matrix}\right.\)