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

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

\(=\frac{3\left(2a+13b\right)-2\left(3a-7b\right)}{3\left(2c+13d\right)-2\left(3c-7d\right)}=\frac{53b}{53d}=\frac{b}{d}\)(1)

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

\(=\frac{7\left(2a+13b\right)+13\left(3a-7b\right)}{7\left(2c+13d\right)+13\left(3c-7d\right)}=\frac{53a}{53c}=\frac{a}{c}\)(2)

Từ (1) (2) => \(\frac{b}{d}=\frac{a}{c}\Rightarrow\frac{c}{d}=\frac{a}{b}\)

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

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

\(\Rightarrow-14ad+39bc=-14cb+39ad\)

\(\Rightarrow-53ad=-53bc\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

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

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

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

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

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

(=) -53ad=-53bc

(=) ad=bc

=> \(\frac{a}{b}=\frac{c}{d}\)

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 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)

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