Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}.C/m:\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

Answer 1

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

Answer 2

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

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

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

\(\Rightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\left(đpcm\right)\)

Answer 3

Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{b}{d}\)

Áp dụng t/c dãy tỉ số bằng nhau :

\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{b}{d}=\dfrac{3a+b}{3c+d}\)

Ta có : \(\dfrac{a}{c}=\dfrac{3a+b}{3c+d}\)

=> \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\) (ĐPCM)

Answer 4

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\left(k\ne0\right)\)

\(\Rightarrow a=bk\) ; \(c=dk\)

Thay \(a=bk\) vào \(\dfrac{a}{3a+b}\) ta có:

\(\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\) (1)

Thay \(c=dk\) vào \(\dfrac{c}{3c+d}\) ta có:

\(\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

Answer 5

đặt \(\dfrac{a}{b}=\(\dfrac{c}{d}=k

=>a=bk ;c=dk

xét \(\dfrac{a}{3a+b}=\(\dfrac{bk}{3bk+b}=\(\dfrac{bk}{b.[3k+1]}=\(\dfrac{k}{3k+1} (1)

xét \(\dfrac{c}{3c+d}=\(\dfrac{dk}{3dk+d}=\(\dfrac{dk}{d.[3k+1]}=\(\dfrac{k}{3k+1} (2)

từ (1) và( 2 ) =>\(\dfrac{a}{3a+b}=\(\dfrac{c}{3c+d}