Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

a: \(\frac{2a+5b}{3a-4b}=\frac{2\cdot bk+5b}{3\cdot bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\frac{2c+5d}{3c-4d}=\frac{2\cdot dk+5d}{3\cdot dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

Do đó: \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

b: \(\frac{3a+7b}{5a-7b}=\frac{3\cdot bk+7b}{5\cdot bk-7b}=\frac{b\left(3k+7\right)}{b\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

\(\frac{3c+7d}{5c-7d}=\frac{3\cdot dk+7d}{5\cdot dk-7d}=\frac{d\left(3k+7\right)}{d\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

Do đó: \(\frac{3a+7b}{5a-7b}=\frac{3c+7d}{5c-7d}\)

d: \(\frac{4a+9b}{4a-7b}=\frac{4\cdot bk+9b}{4\cdot bk-7b}=\frac{b\left(4k+9\right)}{b\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

\(\frac{4c+9d}{4c-7d}=\frac{4\cdot dk+9d}{4\cdot dk-7d}=\frac{d\left(4k+9\right)}{d\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

Do đó: \(\frac{4a+9b}{4a-7b}=\frac{4c+9d}{4c-7d}\)

a/ Đặt :

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

Ta có :

\(\dfrac{2a+7b}{3a-4b}=\dfrac{2bk+7b}{3bk-4b}=\dfrac{b\left(2k+7\right)}{b\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\left(1\right)\)

\(\dfrac{2c+7d}{3c-4d}=\dfrac{2dk+7d}{3dk-4d}=\dfrac{d\left(2k+7\right)}{d\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\)\(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ tương tự

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{5a-7b}{3a+4b}=\frac{5\cdot bk-7b}{3\cdot bk+4b}=\frac{b\left(5k-7\right)}{b\left(3k+4\right)}=\frac{5k-7}{3k+4}\)

\(\frac{5c-7d}{3c+4d}=\frac{5\cdot dk-7d}{3\cdot dk+4d}=\frac{d\left(5k-7\right)}{d\left(3k+4\right)}=\frac{5k-7}{3k+4}\)

Do đó: \(\frac{5a-7b}{3a+4b}=\frac{5c-7d}{3c+4d}\)

Ta đặt:\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

            \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)

giúp mình với, mai mình kiểm tra cuối kỉ rồi

Ta có: \(\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)

\(\Rightarrow\frac{3a+4b}{3a-4b}-1=\frac{3c+4d}{3c-4d}-1\)

\(\Leftrightarrow\frac{8b}{3a-4b}=\frac{8d}{3c-4d}\)

\(\Rightarrow b\left(3c-4d\right)=d\left(3a-4b\right)\)

\(\Leftrightarrow3bc=3ad\)

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

Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow a=bk,c=dk\)

\(\frac{3a+7b}{3c-7d}=\frac{3bk+7b}{3dk+7d}=\frac{b\left(3k+7\right)}{d\left(3k+7\right)}=\frac{b}{d}\)(1)

\(\frac{3a-7b}{3c-7d}=\frac{3bk-7b}{3dk-7d}=\frac{b\left(3k-7\right)}{d\left(3k-7\right)}=\frac{b}{d}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{3a+7b}{3c+7d}=\frac{3a-7b}{3c-7d}\)

1 tỉ số ko thành 1 tỉ lệ thức nhé!