Câu hỏi 14
Rút gọn
$$\frac{(1-\frac1{2^2})(1-\frac1{3^2})(1-\frac1{4^2})\cdots(1-\frac1{2005^2})}{(1-\frac12)(1-\frac13)(1-\frac14)\cdots(1-\frac1{2005})}.$$
$$\frac{(1-\frac1{2^2})(1-\frac1{3^2})(1-\frac1{4^2})\cdots(1-\frac1{2005^2})}{(1-\frac12)(1-\frac13)(1-\frac14)\cdots(1-\frac1{2005})}.$$