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

hhh ngu vc thế mà k bt anh ạ

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

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

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

=>ad-bc=0

=>ad=bc

=>a/b=c/d

=>(a+b)/b=(c+d)/d

Câu trả lời rõ ở link này : https://olm.vn/hoi-dap/detail/9631580415.html

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Suy ra : \(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b.\left(2k+13\right)}{b.\left(3k-7\right)}=\frac{2k+13}{3k-7}\)

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

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

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

<=> (2a+13b)(3c-7d)=(2c+13d)(7a-7b)

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

<=>39bc-14ab=39ab-14bc

<=> bc=ab

<=>\(\frac{a}{b}=\frac{c}{d}\)

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

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

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

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

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

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