Question

[_Girl_Nhân_Mã_]

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

[Huỳnh Quang Sang]

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

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

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

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

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

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

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

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

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

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

Answer 2

[T.Ps]

#)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~#