\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)

Vậy \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{2}\)

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}

\(\frac{1}{51}

Đề là gì z????????????                                                                                        

đây là j`? đầu đề hổng có, làm sao mà giải đc?????

khó vậy 

Ta có : \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

             \(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

              \(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

             \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

              \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

              \(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B=\frac{2015}{51}+\frac{2015}{52}+...+\frac{2015}{100}\)

    \(=2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

\(\Rightarrow\) \(\frac{B}{A}=\frac{2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2015\)

\(\Rightarrow\) \(B⋮A\)

S=1/50+1/51+...+1/98+1/99

Ta thấy :

1/50>1/100

1/51>1/100

................

1/99>1/100

1/100=1/100

=>1/51+1/52+1/53+...+1/98+1/99>1/100+1/100+1/100+...+1/100  (Mỗi bên 50 số hạng)

=>S>50.1/100

=>S>50/100=1/2

Vậy S>1/2

 S lớn hon 1/2 bạn nhé