Giải
a) Từ giả thiết: \(\dfrac{AM}{MB}=\dfrac{7}{4}\Rightarrow\) \(\dfrac{\left(AM+MB\right)}{AM}=\dfrac{\left(7+4\right)}{7}=\dfrac{11}{7}\)
hay \(\dfrac{AB}{AM}=\dfrac{11}{7}:\dfrac{AM}{MB}=\dfrac{7}{4}\)
\(\Rightarrow\dfrac{AM+MB}{MB}=\dfrac{7+4}{4}=\dfrac{11}{4}\) hay \(\dfrac{AB}{BM}=\dfrac{11}{4}\)
b) Ta có: CB = AB - CA = 6cm - 3,6cm = 2,4cm
DA = AB + BD = 6 + BD
Từ giả thiết: \(\dfrac{DA}{DB}=\dfrac{CA}{CB}=\dfrac{3.6}{2.4}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{\left(DB+6\right)}{DB}=\dfrac{3}{2}\)
\(\Rightarrow\) 2DB + 12 = 3DB \(\Rightarrow\) DB = 12 cm