Đặt a b = c d = k ( k ∈ R ) ⇒ a = k . b ; c = k . d
Ta có: 5 a + 3 b 3 a − 7 b = 5 k b + 3 b 3 k b − 7 b = b 5 k + 3 b 3 k − 7 = 5 k + 3 3 k − 7 ( 1 ) 5 c + 3 d 3 c − 7 d = 5 k d + 3 d 3 k d − 7 d = d 5 k + 3 d 3 k − 7 = 5 k + 3 3 k − 7 ( 2 )
Từ (1), (2) => đpcm