Question

cho a, b ,c ,d thoa man \(^{b^2=ac}\)\(^{c^2=bd}\)chung minh \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)

Answer 1

Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)

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

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

\(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Vậy ............................

\(\rightarrowđpcm\)