Do S = \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}\)

\(\Rightarrow\)S = \(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=(1+1+1+...+1) - \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(\Rightarrow\)S=49-\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Dễ thấy:\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)không phải là số tự nhiên

\(\Rightarrow\)S\(\notin N\)

17/25 nha bạn

B  \(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{50^2-1}{50^2}\)

    \(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)

mà    \(0

8/9 x 15/16 x 24/25 x...x 2499/2500

=   \(\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{4\times6}{5\times5}\times...\times\frac{49\times51}{50\times50}\)

=   \(\frac{2\times4\times3\times5\times4\times6\times...\times49\times51}{3\times3\times4\times4\times5\times5\times...\times50\times50}\)

= \(\frac{2\times51}{3\times50}\)

= \(\frac{17}{25}\)

\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{2499}{2500}=\frac{8.15...2499}{9.16...2500}=\frac{2.4.3.5...49.51}{3.3.4.4...50.50}=\frac{\left(2.3.4...49.50\right).\left(4.5...49.51\right)}{\left(2.3.4...49.50\right).\left(3.4.5...49.50\right)}=\frac{51}{3.50}\)\(=\frac{17}{50}\)

HELLO, KẾT BẠN VỚI MÌNH NHÉ ^.^ !!!