Question

\(cho\)cho:\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}.\) cmr: \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)

Answer 1

Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a+b+c}{b+c+d}\)

\(\dfrac{b}{c}=\dfrac{a+b+c}{b+c+d}\)

\(\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\rightarrowđpcm\)

Answer 2

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(1\right)\)

Từ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Ta xét tích: \(\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)=\dfrac{a}{d}\left(dpcm\right)\)