Ta có thể chứng minh :  

Ta có:  

2a+13/b3a−7b=2c+13d/3c−7d

=> 2a+13b/2c+13d=3a−7b/3c−7d

 Áp dụng tính chất của dãy tỉ số bằng nhau ta có :  

2a+13b/2c+13d=3a−7b/3c−7d=2a+13b+3a−7b/2c+13d+3c−7d=5a+6b5c+6d  

Từ 5a+6b/5c+6d = > 5a/5c=6b/6d  

<=> a/c=b/d  

Hay: a/b=c/d (đpcm)

hình như sai rồi

vào học 24 xem bạn ạ

Ta co : \(\frac{2a+13b}{3a-7c}=\frac{2c+13d}{3a-7d}\)

\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a+13b+3a-7b}{2c+13d+3c-7d}=\frac{5a+6b}{5c+6d}\)

Suy ra : \(\frac{5a+6b}{5c+6d}\Rightarrow\frac{5a}{5c}=\frac{6b}{6d}\)

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

Vay : \(\frac{a}{b}=\frac{c}{d}\left(dpcm\right)\)

\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)=>(2a+13b)(3c-7d)=(3a-7b)(2c+13d)

=>6ac-14ad+39bc-91bd=6ac+39ad-14bc-91bd

=>-14ad+39bc=-14bc+39ad

=>-14ad+14bc=39ad-39bc

=>-14(ad-bc)=39(ad-bc)

@-@ sao lại tek này xem lại nhá

2a+13b3a−7b=2c+13d3c−7d→2a+13b2c+13d=3a−7b3c−7d2a+13b3a−7b=2c+13d3c−7d→2a+13b2c+13d=3a−7b3c−7d 

• 2a+13b2c+13d=3a−7b3c−7d=6a+39b6c+39d=6a−14b6c−14d=53b53d=bd2a+13b2c+13d=3a−7b3c−7d=6a+39b6c+39d=6a−14b6c−14d=53b53d=bd (1)

• 2a+13b2c+13d=3a−7b3c−7d=14a+91b14c+91d=39a−91b39c−91d=53a53c=ac2a+13b2c+13d=3a−7b3c−7d=14a+91b14c+91d=39a−91b39c−91d=53a53c=ac (2)

Từ (1) và (2) suy ra ac=bd(=2a+13b2c+13d)→ab=cd

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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b\left(2k+13\right)}{b\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(1\right)\)

\(\frac{2c+13d}{3c-7d}=\frac{2dk+13d}{3dk-7d}=\frac{d\left(2k+13\right)}{d\left(3k-7\right)}=\frac{2k+13}{3k-7}\left(2\right)\)

Từ \(\left(1\right)\) và (2) \(\Rightarrow\frac{a}{b}=\frac{c}{d}\)( đpcm ) 

Chúc bạn học tốt !!!

Từ \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)\(\Rightarrow\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}=\frac{2a}{2c}=\frac{13b}{13d}=\frac{3a}{3c}=\frac{7b}{7d}=\frac{a}{c}=\frac{b}{d}\)

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