Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\begin{cases}a=c.k\\b=d.k\end{cases}\)
Ta có:
\(\frac{7a^2+5ac}{7a^2-5ac}=\frac{a.\left(7a+5c\right)}{a.\left(7a-5c\right)}=\frac{7.c.k+5c}{7.c.k-5c}=\frac{c.\left(7.k+5\right)}{c.\left(7.k-5\right)}=\frac{7.k+5}{7.k-5}\left(1\right)\)
\(\frac{7b^2+5bd}{7b^2-5bd}=\frac{b.\left(7b+5d\right)}{b.\left(7b-5d\right)}=\frac{7.d.k+5d}{7.d.k-5d}=\frac{d.\left(7.k+5\right)}{d.\left(7.k-5\right)}=\frac{7.k+5}{7.k-5}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{7a^2+5ac}{7a^2-5ac}=\frac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)
