\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=t\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)
Ta có đpcm
Ta có :
\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(3\right)\)
Lại có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(4\right)\)
Từ \(\left(3\right)+\left(4\right)\Leftrightarrowđpcm\)
