viết có chắc chữ giải mà cũng đúng thật vô lý

a)

\(\left(1+\frac{1}{51}\right).\left(1+\frac{1}{52}\right).\left(1+\frac{1}{53}\right)...\left(1+\frac{1}{1963}\right)\)

\(=\frac{52}{51}.\frac{53}{52}.\frac{54}{53}...\frac{1964}{1963}\)

\(=\frac{1964}{51}\)

b)

\(\left(1+\frac{1}{400}\right).\left(1+\frac{1}{401}\right).\left(1+\frac{1}{402}\right)...\left(1+\frac{1}{2010}\right)\)

\(=\frac{401}{400}.\frac{402}{401}.\frac{403}{402}...\frac{2011}{2010}\)

\(=\frac{2011}{400}\)

1.3.5.7........99 = \(\frac{\left(1.3.5.7......99\right)\left(2.4.6......100\right)}{2.4.6......100}\)= \(\frac{1.2.3......99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.......100}{2.2.2......2}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\)(ĐPCM)

                                      50 số 2

ko hiểu

la sao bạn

Ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

@Ác Mộng ở đoạn cuối tự nhiên bỏ mất số 2 luôn, giải sai rồi kìa