Ta có: a. (c + d) = a.c + a.d
c. (a + b) = c.a +c.b
Vì \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) ( đề cho) => a.d = b.c
=> a.c + a.d = c.a +c.b
=> a.( c+ d) = (a + b).c
=> \(\dfrac{a}{a+b}\) = \(\dfrac{c}{c+d}\) ( đpcm)
Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\Rightarrow\dfrac{b}{a}+1=\dfrac{d}{c}+1\Rightarrow\dfrac{b+a}{a}=\dfrac{d+c}{c}\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{d+c}\)
Vậy ....
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
Ta có:
\(1:\dfrac{a}{a+b}=\dfrac{a+b}{a}=1+\dfrac{b}{a}\)
\(1:\dfrac{c}{c+d}=\dfrac{c+d}{c}=1+\dfrac{d}{c}\)
Vì \(\dfrac{b}{a}=\dfrac{d}{c}\) nên \(1+\dfrac{b}{a}=1+\dfrac{d}{c}\). \(\Rightarrow\) \(1:\dfrac{a}{a+b}=1:\dfrac{c}{c+d}\). \(\Rightarrow\) \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(\Rightarrow\) ĐPCM