Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(đpcm\right)\)
Ta đặt: k = \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\)
=>k3= \(\dfrac{a}{b}\).\(\dfrac{b}{c}\).\(\dfrac{c}{d}\)=\(\dfrac{a}{d}\) (1)
Lại có: k = \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\) (2)
Từ (1), (2) =>( \(\dfrac{a+b+c}{b+c+d}\))3 = k3= \(\dfrac{a}{d}\)