-Lớp 5 làm gì biết dấu nhân trong \(AM=\dfrac{1}{4}AB\) được ẩn?
a. -Vì \(AN=\dfrac{1}{4}AC\) và \(AN+NC=AC\).
Nên \(\dfrac{1}{4}AC+NC=AC\)
\(NC=AC-\dfrac{1}{4}AC=AC\times\left(1-\dfrac{1}{4}\right)=\dfrac{3}{4}AC\).
\(\dfrac{S_{BNC}}{S_{ABC}}=\dfrac{NC}{AC}=\dfrac{\dfrac{3}{4}AC}{AC}=\dfrac{3}{4}\)
\(\dfrac{S_{BNC}}{64}=\dfrac{3}{4}\)
\(S_{BNC}=\dfrac{3}{4}\times64=48\left(cm^2\right)\).
b. \(\dfrac{S_{AMN}}{S_{ANB}}=\dfrac{AM}{AB}=\dfrac{\dfrac{1}{4}AB}{AB}=\dfrac{1}{4}\)
\(\dfrac{S_{ANB}}{S_{ABC}}=\dfrac{AN}{AC}=\dfrac{\dfrac{1}{4}AC}{AC}=\dfrac{1}{4}\)
\(\dfrac{S_{AMN}}{S_{ANB}}\times\dfrac{S_{ANB}}{S_{ABC}}=\dfrac{1}{4}\times\dfrac{1}{4}\)
\(\dfrac{S_{AMN}}{S_{ABC}}=\dfrac{1}{16}\)