Ta có: `x/3 = y/5 = k`
`⇒ x = 3k` ; `y = 5k`
Khi đó: `A = (5x^2 + 3y^2)/(5x^2 - y^2) = (5(3k)^2 + 3(5k)^2)/(5(3k)^2 + (5k)^2) = (5. 9k^2 + 3. 25k^2)/(5. 9k^2 + 25k^2) = (45k^2 + 75k^2)/(45k^2 + 25k^2) = (k^2(45 + 75))/(k^2(45 + 25)) = 120/70 = 12/7`
Vậy `A = 12/7` khi `x/3 = y/5`