Ta có:
\(c.\left(b+d\right)=2bd\)
\(\Rightarrow bc+cd=2bd\)
Lại có: \(a+c=2b\)
Lấy vế chia vế được: \(\dfrac{bc+cd}{a+c}=\dfrac{2bd}{2b}=d\)
\(\Rightarrow bc+cd=ad+cd\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
* \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(\dfrac{a+c}{b+d}\right)^8=\left(\dfrac{a}{b}\right)^8=\dfrac{a^8}{b^8}\left(1\right)\)
* \(\dfrac{a}{b}=\dfrac{c}{d}=\left(\dfrac{a}{b}\right)^8=\left(\dfrac{c}{d}\right)^8\)
\(\Rightarrow\dfrac{a^8}{b^8}=\dfrac{c^8}{d^8}=\dfrac{a^8+c^8}{b^8+d^8}\left(2\right)\)
Từ (1) và (2) suy ra:
\(\left(\dfrac{a+c}{b+d}\right)^8=\dfrac{a^8+c^8}{b^8+d^8}\left(đpcm\right)\)
