Question

Tính:

\(P=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}}\)

Answer 1

cô trang dạy rồi mà

Answer 2

Khó kinh .."

Answer 3

Gọi \(Q=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

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

\(\Rightarrow Q=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrow Q=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-2\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

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

\(\Rightarrow P=1\)

Answer 4

từ dòng thứ tư sửa thành Q=(1+1/2+1/3+...+1/99)-2(1/2+1/4+...+1/100) mới ra đc Q=1/51+1/52+...+1/100

Answer 5

chả hiểu sao cho đúng đc