Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\frac{3a^2+5ab}{7a^2-10b^2}=\frac{3.\left(bk\right)^2+5.bkb}{7\left(bk\right)^2-10b^2}=\frac{3b^2k^2+5kb^2}{7b^2k^2-10b^2}=\frac{kb^2\left(3k+5\right)}{b^2\left(7k^2-10\right)}=\frac{k\left(3k+5\right)}{\left(7k^2-10\right)}\left(1\right)\)
\(\frac{3c^2+5cd}{7c^2-10d^2}=\frac{3.\left(dk\right)^2+5dkd}{7\left(dk\right)^2-10d^2}=\frac{3d^2k^2+5kd^2}{7d^2k^2-10d^2}=\frac{kd^2\left(3k+5\right)}{d^2\left(7k^2-10\right)}=\frac{k\left(3k+5\right)}{\left(7k^2-10\right)}\left(2\right)\)
Từ (1) và (2)
⇒ĐPCM
