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![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(\frac{a}{b}\)= \(\frac{c}{d}=k\Rightarrow\hept{\begin{cases}k=ab\\k=cd\end{cases}}\)
ta có : \(\frac{7a-4b}{3a+5b}\)= \(\frac{7ak-4b}{3ak-5b}=\frac{a\left(7k-4\right)}{a\left(3k-5\right)}=\frac{7k-4}{3k-5}\left(1\right)\)
\(\frac{7c-4d}{3c+5d}\)=\(\frac{7ck-4d}{3ck+5d}\)= \(\frac{c\left(7k-4\right)}{c\left(3k+5\right)}\)= \(\frac{7k-4}{3k+5}\)( 2 )
từ (1) và ( 2) => \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)( điều phải chứng minh )
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)