Ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+c}{b+d}=\frac{a+2c}{a+2d}\Leftrightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\)

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

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

Theo TCDTSBN:

\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)

Áp dụng TCDTSBN ta có:

\(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{2c}{2d}=\frac{4c}{4d}=\frac{5a+2c}{5b+2d}=\frac{a-4c}{b-4d}\)

k nhé!