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

\(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\Rightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\) (1)

Nhân tư và mẫu vế trái (1) với 3 và vế phải với 13 ta được:

\(\dfrac{2a+13b}{2c+13d}=\dfrac{14a+91b}{14c+91d}=\dfrac{39a-91b}{39c-91d}\)

=\(\dfrac{\left(14a+91b\right)+\left(39a-91b\right)}{\left(14c+91d\right)+\left(39c-91d\right)}=\dfrac{53a}{53c}=\dfrac{a}{c}\) (2)

Nhân tử và mẫu vế trái (1) với 3 và vế phải với 2 ta được:

\(\dfrac{2a+13b}{2c+13d}=\dfrac{6a+39b}{6c+39d}=\dfrac{6a-14b}{6c-14d}=\dfrac{53b}{53d}=\dfrac{b}{d}\) (3)

Từ (2) và (3) suy ra :

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

bạn ấy giải chỉ nhầm một tẹo

bạn nào vậy bạn mà sai mootjtis cũng là sai rồi

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

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:

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