Trước hết ta đi chứng minh 1 bài toán \(a^2-1=\left(a-1\right)\left(a+1\right)\) (1)
Thật vậy \(\left(a-1\right)\left(a+1\right)=\left(a-1\right)a+\left(a-1\right)=a^2-a+a-1=a^2-1\)
Áp dụng (1) ta có:
\(A=\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{7\cdot9}{8^2}\\ =\dfrac{2\cdot3\cdot4^2\cdot...\cdot7^2\cdot8\cdot9}{3^2\cdot4^2\cdot...\cdot8^2}=\dfrac{2\cdot9}{3\cdot8}\\ =\dfrac{18}{24}=\dfrac{3}{4}\)