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

\(\Rightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)

\(\Rightarrow6ac+39bc-14ad-91bd=6ac+39ad-14bc-91bd\)

\(\Rightarrow6ac-6ac+39bc+14bc-14ad-39ad-91bd+91bd=0\)

\(\Rightarrow53bc-53ad=0\)

\(\Rightarrow53bc=53ad\)

\(\Rightarrow bc=ad\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm.\)

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

\(\Leftrightarrow\)(2a+13b)(3c-7d)=(2c+13d)(3a-7b)

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

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

14ad+39bc+91bd=14bc+39ad+91bd

14ad+39bc=14bc+39ad

39bc=14bc+39ad-14ad

39bc=14bc+25ad

39bc-14bc=25ad

25bc=25ad

bc=ad

Ta có: Điều đề bài cho:

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\left(đpcm\right)\)

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

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

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

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

hay \(\dfrac{a}{b}=\dfrac{c}{d}\)

Nguyễn Huy Tú chắc làm sai rồi

Chứng minh:

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

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

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

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

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left\{{}\begin{matrix}a=b\\c=d\end{matrix}\right.\Rightarrow\dfrac{a}{a}=\dfrac{c}{c}\)

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

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

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}=\dfrac{2a}{2c}=\dfrac{13b}{13d}=\dfrac{3a}{3c}=\dfrac{7b}{7d}=\dfrac{a}{c}=\dfrac{b}{d}\)

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

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

Vậy \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

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

\(\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)

\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)

\(\Leftrightarrow-53ad=-53bc\)

=>ad=bc

hay a/b=c/d

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

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

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