Theo tính chất dãy tỉ số ta có:
\(\dfrac{a}{b}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)
\(\Rightarrow\)\(\dfrac{a-c}{a+c}\)=\(\dfrac{c-b}{c+b}\)(đpcm)
\(\Rightarrow\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{a+c}{c+b}=\dfrac{a-c}{c-b}\Rightarrow\dfrac{a+c}{a-c}=\dfrac{c+b}{c-b}\Rightarrow\dfrac{a-c}{a+b}=\dfrac{c-b}{c+b}\)
Cho :\(\dfrac{a}{c}=\dfrac{c}{b}\)
CMR: \(\dfrac{a-c}{a+c}=\dfrac{c-b}{c+b}\)
Ta có: \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{a+c}{c+b}=^{\dfrac{ }{ }}\)\(\dfrac{a-c}{c-b}\)
(Cái này bn suy theo tỉ lệ thức nhé)
=> \(\dfrac{a+c}{a-c}=\dfrac{c+b}{c-b}\)
*Tích cho mik với nhé!
- Đặt \(\dfrac{a}{c}\) = \(\dfrac{c}{b}\) = k
=> \(\dfrac{a}{c}\)= k => a = k.c
\(\dfrac{c}{b}\)= k => c = k.b
Ta có: \(\dfrac{a-c}{a+c}\)= \(\dfrac{k.c-k.b}{k.c+k.b}\)= \(\dfrac{k.\left(c-b\right)}{k.\left(c+b\right)}\)= \(\dfrac{c-b}{c+b}\)
Vì \(\dfrac{c-b}{c+b}\)=\(\dfrac{c-b}{c+b}\)hay\(\dfrac{a-c}{a+c}\) = \(\dfrac{c-b}{c+b}\)
Vậy \(\dfrac{a-c}{a+c}\)= \(\dfrac{c-b}{c+b}\)( Đpcm )
