Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)

\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)

Do đó: \(\dfrac{3a+2b}{a}=\dfrac{3c+2d}{c}\)

b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)

\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)

Do đó: \(\dfrac{2a-3b}{b}=\dfrac{2c-3d}{d}\)

c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)

\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)

Do đó: \(\dfrac{a}{a-2b}=\dfrac{c}{c-2d}\)

a/b = c/d => a/c = b/d => (a+2c)/(a+c) = (b+2d)/(b+d) 
=> (a+2c)(b+d) = (a+c)(b+2d) 
thế này đúng ko

Áp dụng tính chất dãy tỉ số bằng nhau:

 a/b = c /d = (a+c )/(b+d)  

a/b = c/d = 2c/2d= (a+2c)/(b+2d)  

=> (a+c )/(b+d)=(a+2c)/(b+2d)  

=> ( a+c)(b+2d)=(b+d)( a+2c) 

**** nhe

Đặt k        

Ta thấy : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow ad=bc\)(1)

Ta có:   (a+2c)(b+d)=(a+c)(b+ad)

<=> ab+ad+2bc+2cd=ab+2ad+bc+2cd

<=> ab+ad+2bc+2cd-ab-2ad-bc-2cd=0

<=>-ad+bc=0<=>bc-ad=0<=>ad=bc=>(1) luôn đúng

=>ĐFCM

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

Khi đó : \(\frac{a+2c}{b+2d}=\frac{bk+2dk}{b+2d}=\frac{k\left(b+2d\right)}{b+2d}=k\left(1\right)\)

\(\frac{a-3c}{b-3d}=\frac{bk-3dk}{b-3d}=\frac{k\left(b-3d\right)}{b-3d}=k\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\left(\text{đpcm}\right)\)