Question

Tính A = ( 1/ 1.2 + 1/3.4 + .... + 1/399.400 ) : ( 1/ 201.400 + 1/202.399 + ... + 1/300.301 )

Answer 1

Đặt C = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{399.400}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\)

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{400}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{400}\right)\)

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{400}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{400}\)(1) 

Đặt B = \(\frac{1}{201.400}+\frac{1}{202.399}+...+\frac{1}{300.301}\)

=> 601B = \(\frac{601}{201.400}+\frac{601}{202.399}+...+\frac{601}{301.300}=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)

=> B = \(\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\right):601\)

Khi đó : \(A=\frac{C}{B}=\frac{\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}}{\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\right):601}=601\)

Vậy A = 601 

Answer 2

thx bạn xyz nha !