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#)Giải :

Ta có : \(c\left(b+d\right)=2bd\Rightarrow bc+cd=2bd\Rightarrow\frac{bc+cd}{a+c}=\frac{2bd}{2b}=d\)

\(\Rightarrow bc+cd=ad+cd\Rightarrow bc=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\hept{\begin{cases}\left(\frac{a}{b}\right)^8=\left(\frac{c}{d}\right)^8=\left(\frac{a+c}{b+d}\right)^8=\frac{a^8}{b^8}\\\left(\frac{a}{b}\right)^8=\left(\frac{c}{d}\right)^8=\frac{a^8}{b^8}=\frac{c^8}{d^8}=\frac{a^8+c^8}{b^8+c^8}\end{cases}}\)

\(\Rightarrow\left(\frac{a+c}{b+d}\right)^8=\frac{a^8+b^8}{c^8+d^8}\left(đpcm\right)\)

Từ c(b+d)=2bd=>bc+cd=2bd

Ta lại có             a+c =2b

Lấy vế chia vế được :\(\frac{bc+cd}{a+c}=\frac{2bd}{2b}=\)\(d\)

=>bc+cd=ad+cd=>bc=ad=>\(\frac{a}{b}=\frac{c}{d}\)

+ , \(\frac{a}{b}=\frac{c}{d}\)= \(\frac{a+c}{b+d}\)=> \(\left(\frac{a+c}{b+d}\right)^8=\left(\frac{a}{b}\right)^8\)= \(\frac{a^8}{b^8}\) (1)

+ \(\frac{a}{b}=\frac{c}{d}\)=> \(\left(\frac{a}{b}\right)^8=\left(\frac{c}{d}\right)^8\)<=>\(\frac{a^8}{b^8}=\frac{c^8}{d^8}\)=\(\frac{a^8+c^8}{b^8+d^8}\) (2)

Từ (1) và (2) ta suy ra : \(\left(\frac{a+c}{b+d}\right)^8=\frac{a^8+c^8}{b^8+d^8}\) ( đpcm)

\(b=\frac{a+c}{2}\Rightarrow2b=a+c\Rightarrow2bd=d\left(a+c\right)=ad+dc\)  (1)

\(c=\frac{2bd}{b+d}\Rightarrow2bd=c\left(b+d\right)=cb+cd\) (2)

Từ (1) và (2) => \(ad+dc=cb+cd\)                   \(\left(=abd\right)\)

=> \(ad+cd-cd=cb+cd-cd\)

=> \(ad=cb\)

=> \(\frac{a}{b}=\frac{c}{d}\)

vậy 4 số a, b, c, d lập đc 1 tỉ lệ thức

Ta có b=\(\dfrac{a+c}{2}\)⇒2b=a+c⇒2bd=d(a+c)=ad+dc(1)

          c=\(\dfrac{2bd}{b+d}\)⇒2bd=c(b+d)=cb+cd(2)

Từ (1) và (2)⇒ad+dc=cb+cd(=2bd)

⇒ad+cd-cd=cb+cd-cd

⇒ad=cb

2bd=c(b+d)

<=>(a+c)d=bc+cd

<=>ad+cd=bc+cd

<=>ad=bc

<=>\(\frac{a}{b}=\frac{c}{d}\)

<=>\(\frac{a}{c}=\frac{b}{d}\) <=> \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)<=>\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) (đpcm)

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)