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 có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:

\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)

\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)

\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)

Cộng (1),(2) và (3) có:

\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)

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

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

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

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

\(\Leftrightarrow3bc=3ad\)

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

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

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