Question

       Giúp mình nha:

 Bài 1 ; Chứng minh rằng:

a, 1/1.2+1/3.4+5.6 + ........+1/99.100=1/51+1/52+1/53+.....+1/100

b, Tính (1/101+ 1/102+.....+1/200): (1/1.2+1/3.4+1/5.6+.....+1/199.200)

 MÌNH CẢM ƠN NHÌU ! 

Bạn nào giúp mình trước mình tick nha.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 1

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

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

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

Answer 2

cảm ơn nguyenthuylinh nha