\(\frac{a}{b}+\frac{c}{d}=\frac{a.c}{b.d}\)

\(vậy:\frac{ad+bc}{bd}=\frac{ac}{bd}\)

        =>\(ad+bd=ac\)

        =>\(ad=ac-bc\)

       =>\(ad=c\left(a-b\right)\)

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

\(\frac{a}{b}:\frac{c}{d}=\frac{a}{b}.\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{a}{b}.\frac{c}{d}.\frac{c}{d}\Rightarrow\frac{a}{b}:\frac{a}{b}=\frac{c}{d}.\frac{c}{d}\)

\(\Rightarrow1=\left(\frac{c}{d}\right)^2\Rightarrow\frac{c}{d}=1\text{ hoặc }\frac{c}{d}=-1\)

ta có: \(\frac{a}{b}:\frac{c}{d}=\frac{a}{b}.\frac{c}{d}\Leftrightarrow\frac{a}{b}.\frac{d}{c}=\frac{a}{b}.\frac{c}{d}\Leftrightarrow\frac{a.d}{b.c}=\frac{a.c}{bd}\Leftrightarrow\frac{d}{c}=\frac{c}{d}\Leftrightarrow d^2=c^2\)

suy ra d=c hoặc d=-c

suy ra \(\frac{c}{d}=\frac{c}{c}=1\) hoặc \(\frac{c}{d}=\frac{c}{-c}=-1\)