\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\\ \Rightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\\ \Rightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\\ \Rightarrow\dfrac{-\left(a-b\right)}{a}=\dfrac{-\left(c-d\right)}{c}\\ \Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

#)Giải :

a) Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ad+bd=bc+bd\Rightarrow d\left(a+b\right)=b\left(c+d\right)\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)

\(\Rightarrowđpcm\)

b) Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ad-bd=bc-bd\Rightarrow d\left(a-b\right)=b\left(c-d\right)\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)

\(\Rightarrowđpcm\)

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)

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

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

\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\left(đpcm\right).\)

Chúc bạn học tốt!

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) (k\(\in\)N*)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Thay vào ta có:

\(\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1}\)

\(\frac{c}{c+d}=\frac{dk}{dk+d}=\frac{dk}{d\left(k+1\right)}=\frac{k}{k+1}\)

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

Vậy \(\frac{a}{a+b}=\frac{c}{c+d}\)(điều phải chứng minh)

Hok tốt nha!!!

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{3a+4b}{5a-3b}=\dfrac{3\cdot bk+4b}{5\cdot bk-3b}=\dfrac{b\left(3k+4\right)}{b\left(5k-3\right)}=\dfrac{3k+4}{5k-3}\)

\(\dfrac{3c+4d}{5c-3d}=\dfrac{3\cdot dk+4d}{5\cdot dk-3d}=\dfrac{d\left(3k+4\right)}{d\left(5k-3\right)}=\dfrac{3k+4}{5k-3}\)

Do đó: \(\dfrac{3a+4b}{5a-3b}=\dfrac{3c+4d}{5c-3d}\)