b, Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k,c=d.k\)
+) \(\frac{7a^2+3ab}{11a^2-8ab^2}=\frac{7k^2.b^2+3kb^2}{11k^2.b^2-8b^2}=\frac{7k^2+3k}{11k^2-8}\left(1\right)\)
+) \(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7k^2.d^2+3k.d^2}{11k^2.d^2-8d^2}=\frac{7k^2+3k}{11k^2-8}\left(2\right)\)
Từ (1) và (2) => ĐPCM