cách 1:
$\frac{2a+13b}{3a-7b}$=$\frac{2c+13d}{3c-7d}$ <=> $\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}$
<=> $(2c+13d)(3a-7b)=(2a+13b)(3c-7d)$
=> $14bc+39ad=14ad+39bc$ (tự nhân thì ra)
=> $14(bc-ad)+39(ad-bc)=0$
Đặt $bc-ad=x$
=> $14x-39x=0$
=> $-25x=0$
=> $x=0$
=> $bc=ad$
=> $\frac{a}{b}=\frac{c}{d}$
Cách 2: ko biết
Làm thế này nè bạn:
Cách 1:
Ta có:$\frac{2a+13b}{3a-7b}$=$\frac{2c+13d}{3c-7d}$
$\Leftrightarrow$ (2a+13b)(3c-7d)=(2c+13d)(3a-7b)
$\Leftrightarrow$ 6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd
$\Leftrightarrow$ 6ac-6ac-14ad+14bc=39ad-39bc-91bd+91bd
$\Leftrightarrow$ -14ad+14bc=39ad-39bc
$\Leftrightarrow$ 14bc+39bc=39ad+14ad
$\Leftrightarrow$ 53bc=53ad
$\Leftrightarrow$ bc=ad
$\Rightarrow$ $\frac{a}{b}=\frac{c}{d}$
Cách 2:
Ta có:$\frac{2a+13b}{3a-7b}$=$\frac{2c+13d}{3c-7d}$
$\Leftrightarrow$ $\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}$ =$\frac{14a+91b}{14c+91b}$=$\frac{39a-91b}{39c-91b}$=$\frac{53a}{53c}$=$\frac{a}{c}$(1)
Mặt khác:
$\frac{2a+13b}{2c+13d}=\frac{3a-7b}{3c-7d}$=$\frac{6a+39b}{6c+39d}$=$\frac{6a-14b}{6a-14d}$=$\frac{53b}{53d}$=$\frac{b}{d}$
Từ (1) và (2) $\Rightarrow$ $\frac{a}{c}=\frac{b}{d}$
hay $\frac{a}{b}=\frac{c}{d}$