Question

Tính các tổng sau đây:

a)A=\(\frac{1}{25.27}\) +\(\frac{1}{27.29}\) +....+\(\frac{1}{73.75}\)

b)B=\(\frac{1}{8.11}\) +\(\frac{1}{11.14}\) +\(\frac{1}{14.17}\) +....+\(\frac{1}{197.200}\)

CÁC BẠN ƠI,GIÚP MIK VỚI.AI NHANH,LÀM ĐÚNG VÀ ĐỦ 2 Ý THÌ MIK TICK CHO NHA.Thank you very much.

Answer 1

\(a,A=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)

\(A=\frac{1}{2}\left[\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+...+\frac{2}{73\cdot75}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{75}\right]=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(b,B=\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+...+\frac{1}{197\cdot200}\)

\(3B=\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\)

\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)

\(3B=\frac{1}{8}-\frac{1}{200}\)

\(3B=\frac{3}{25}\)

\(B=\frac{3}{25}:3=\frac{1}{25}\)

Answer 2

#)Giải :

a, \(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(A=\frac{1}{25}-\frac{1}{75}\)

\(A=\frac{2}{75}\)

b, \(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

\(B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

\(B=\frac{1}{8}-\frac{1}{200}\)

\(B=\frac{3}{25}\)

            #~Will~be~Pens~#

Answer 3

\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2A=\frac{1}{25}-\frac{1}{75}\)

\(2A=\frac{2}{75}\)

\(A=\frac{2}{75}:2=\frac{1}{75}\)

\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)

\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

\(3B=\frac{1}{8}-\frac{1}{200}\)

\(3B=\frac{3}{25}\)

\(B=\frac{3}{25}:3=\frac{1}{25}\)

Answer 4

#)Ui chết sorry bn, mk ghi thiếu nhé @@

  ph là 2A và 3B ms đúng :v

  Mấy bn kia lm đúng rùi ! 

Answer 5

a) \(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(\Leftrightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\Rightarrow A=\frac{1}{75}.\)

b)\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)

\(\Leftrightarrow3B=\frac{1}{8}-\frac{1}{200}=\frac{3}{25}\Leftrightarrow B=\frac{1}{25}.\)

Answer 6

A = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\)\(\frac{1}{73.75}\)

2A = \(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

2A = \(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

2A = \(\frac{1}{25}-\frac{1}{75}\)

2A = \(\frac{3}{75}-\frac{1}{75}\)

=>A = \(\frac{2}{75}:2\)=\(\frac{2}{75}.\frac{1}{2}\)

=> A =\(\frac{1}{75}\)

B = \(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

3B = \(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)

3B = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

3B = \(\frac{1}{8}-\frac{1}{200}\)

=> 3B = \(\frac{25}{200}-\frac{1}{200}\)

=> B = \(\frac{24}{200}.\frac{1}{3}\)

=> B = \(\frac{1}{25}\)

#Học'Tốt#

Answer 7

\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2A=\frac{1}{25}+\left(\frac{1}{27}-\frac{1}{27}\right)+\left(\frac{1}{29}-\frac{1}{29}\right)+...+\left(\frac{1}{73}-\frac{1}{73}\right)-\frac{1}{75}\)

\(2A=\frac{1}{25}-\frac{1}{75}\)

\(2A=\frac{2}{75}\)

\(A=\frac{2}{75}:2\)

\(A=\frac{1}{75}\)

~ Hok tốt ~

Answer 8

\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)

\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)

\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)

\(3B=\frac{1}{8}+\left(\frac{1}{11}-\frac{1}{11}\right)+\left(\frac{1}{14}-\frac{1}{14}\right)+...+\left(\frac{1}{197}-\frac{1}{197}\right)-\frac{1}{200}\)

\(3B=\frac{1}{8}-\frac{1}{200}\)

\(3B=\frac{3}{25}\)

\(B=\frac{3}{25}:3\)

\(B=\frac{1}{25}\)

~ Hok tốt ~