Ta có :

\(c=\frac{bd}{b-d}\)

\(\Rightarrow b-d=\frac{bd}{c}\left(c\ne0\right)\)

\(a=b+c\Rightarrow c=a-b\)

\(\Rightarrow c=\frac{bd}{b-d}=a-b\)

\(\Rightarrow bd=\left(a-b\right).\left(b-d\right)\)

\(\Rightarrow ab-ad-b^2+bd=bd\)

\(\Rightarrow a\left(b-d\right)-b^2=0\)

\(\Rightarrow a.\frac{bd}{c}-b^2=0\)

\(\Rightarrow\frac{ad}{c}-b=0\)

\(\Rightarrow\frac{ad-bc}{c}=0\)

\(\Rightarrow ad-bc=0\)

\(\Rightarrow ad=bc\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

Chúc bạn học tốt !!!

\(c=\frac{bd}{b-d}\)

=> c(b - d) = bd

=> bc - cd = bd

=> bc = bd + cd

=> bc = d(b + c)

=> bc = ad

=> \(\frac{a}{b}=\frac{c}{d}\)

cảm ơn bạn nhiều nhiều !!!!!!!!!!!!!!!!!

Ta có:

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)

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

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

Ta có : \(b^2=ac\) 

\(\Rightarrow\frac{a}{b}=\frac{b}{c}\) (1) 

\(c^2=bd\) 

\(\Rightarrow\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1) và (2) suy ra : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) 

\(\Rightarrow\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) , \(\frac{b}{c}.\frac{b}{c}.\frac{b}{c}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\) và \(\frac{c}{d}.\frac{c}{d}.\frac{c}{d}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)

\(\Rightarrow\frac{a^3}{b^3}=\frac{a}{d}\) , \(\frac{b^3}{c^3}=\frac{a}{d}\) và \(\frac{c^3}{d^3}=\frac{a}{d}\) 

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

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

Vậy \(\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

 Ta có:

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)

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

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

\(ADTCDTSBN,\)ta có:

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

Lại có:\(\frac{a^3}{b^3}=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)

Ta có : \(\hept{\begin{cases}b^2=ac\\c^2=bd\end{cases}\Rightarrow\hept{\begin{cases}b.b=a.c\\c.c=b.d\end{cases}\Rightarrow}\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}\\\frac{b}{c}=\frac{c}{d}\end{cases}\Rightarrow}\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}}\)

=> \(\frac{a^3}{b^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)(1)

mà \(\frac{a^3}{b^3}=\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{abc}{bcd}=\frac{a}{d}\)(2) 

Từ (1) và (2) => đpcm 

\(c=\frac{bd}{b-d}\Rightarrow bc-dc=bd\Rightarrow bc=bd+dc=d\left(b+c\right)\)

Mà \(a=b+c\)nên\(bc=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}=\frac{2c}{2d}=\frac{5a}{5b}\)

ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU 

\(\frac{2a+2c}{2b+2d}=\frac{5a-c}{5b-d}\)

MÌNH SỬA LẠI ĐỀ LÀ 3D THÀNH 2D NHÉ