Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)
\(\frac{a+b}{b}=\frac{kb+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\)(1)
\(\frac{c+d}{d}=\frac{kd+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\)(2)
Từ (1) và (2) => \(\frac{a+b}{b}=\frac{c+d}{d}\)=> đpcm
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a+b}{b}=\frac{c+d}{d}\Rightarrow min=\frac{1+2}{2}=\frac{2+4}{4}=\frac{3}{2}=\frac{6}{4}\)
\(\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)