Ta co:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\)

               \(=\frac{2007b}{2007c}=\frac{a+2007b}{b+2007c}\)

     \(\Rightarrow\left(\frac{a+2007b}{b+2007c}\right)^2=\left(\frac{a}{b}\right)^2=\frac{a}{b}\times\frac{b}{c}=\frac{a}{c}\)

          Vậy \(\frac{a}{c}=\left(\frac{a+2007b}{b+2007c}\right)^2\left(đpcm\right)\)

b² = ac => \(\frac{a}{b}=\frac{b}{c}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{a+2007b}{b+2007c}\)

=> \(\left(\frac{a+2007b}{b+2007c}\right)^2=\frac{a+2007b}{b+2007c}.\frac{a+2007b}{b+2007c}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)

Vậy \(\frac{a}{c}=\left(\frac{a+2007b}{b+2007c}\right)^2\)

Ta có:  \(b^2=a.c\Rightarrow\frac{a}{b}=\frac{b}{c}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\left(k\in R\right)\)

\(\Rightarrow a=b.k\); \(b=c.k\)

\(\frac{a}{c}=\frac{a.c}{c.c}=\frac{b^2}{c^2}\left(1\right)\)

\(\frac{\left(a+2007b\right)^2}{\left(b+2007c\right)^2}=\frac{\left(b.k+2007b\right)^2}{\left(c.k+2007c\right)^2}=\frac{\left[b\left(k+2007\right)\right]^2}{\left[c.\left(k+2007\right)\right]^2}=\frac{b^2.\left(k+2007\right)^2}{c^2.\left(k+2007\right)^2}=\frac{b^2}{c^2}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{a}{c}=\frac{\left(a+2007b\right)^2}{\left(b+2007c\right)^2}\) \(\left(đpcm\right)\) 

\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng t/c dtsbn:

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

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

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)