a ) \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)

\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{203}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{203}\right)\)

\(=\frac{3}{2}.\left(\frac{203}{1015}-\frac{5}{1015}\right)\)

\(=\frac{3}{2}.\frac{198}{1015}\)

\(=\frac{297}{1015}\)

b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Tham khảo  nha !!! 

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{201\cdot203}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{201}-\frac{1}{203}\)

\(2A=1-\frac{1}{203}\)

\(A=\frac{101}{203}\)

Đáp án là : \(\frac{101}{203}\)

100% là 101 / 203 luôn

6A=1.3.6 + 3.5.6 + 5.7.6 + ... + 201.203.6

6A=1.3(5+1) + 3.5(7-1) + 5.7(9-3) + ... + 201.203(205-199)

6A=(1.3.5 + 1.3) +( 3.5.7 - 1.3.5) + (5.7.9 - 3.5.7) + ... + (201.203.205 - 199.201.203)

6A=3+201.203.205

6A=8364618

A=\(\frac{8364618}{6}\)

A=1394103

máy tính cũng phải hỏng với phét tính này

\(\dfrac{2^3}{3\cdot5}+\dfrac{2^3}{5\cdot7}+...+\dfrac{2^3}{101\cdot103}\)

\(=2^2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{101\cdot103}\right)\)

\(=4\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)

\(=4\cdot\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=4\cdot\dfrac{100}{309}=\dfrac{400}{309}\)

\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)

\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)

\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(=4\cdot\frac{100}{309}=\frac{400}{309}\)

a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)

b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)