$\frac{7a^2+ 3ab}{11a^2 -8b^2}$ = $\frac{7c^2 + 3cd}{11c^2-8d^2}$
Đặt
$\frac{a}{b}$ = $\frac{c}{d}$=k => a= bk; c= dk
Ta có:
VT(1): $\frac{7a^2 + 3ab}{11a^2-8b^2}$ =$\frac{7bk^2+3bk.b}{11bk^2-8b^2}$ =$\frac{b^2(7k^2+ 3k^2}{b^2(11k^2 -8k2}$ =$\frac{7k^2 + 3k}{11a^2 -8}$ (1')
VP(1):
$\frac{7c^2+3cd}{11c^2-8cd}$ =$\frac{7dk^2+3dk.d}{11dk^2-8d^2}$ =$\frac{d^2(7k^2+3k^2)}{D^2(11k^2-8k^2}$ =$\frac{7k^2+3k}{11k^2-8}$(1")
Từ (1') (1")=> VT(1)=VP(1) (đpcm)
