a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99

Tổng trên có số số hạng là:

(99 - 3) : 4 + 1 = 25 (số)

Tổng trên là:

(99 + 3) x 25 : 2 = 1275

ĐS:

A=[99-1]:2+1

=49

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

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

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

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

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}\)

\(B=\frac{12}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(C=7.\frac{3}{35}\)

\(C=\frac{3}{5}\)

Ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)

\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\) 

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}\)

\(\Rightarrow C=\frac{3}{5}\)

(4+11):15+(35+64):99+(2+5):7=15:15+99:99+7:7=1+1+1=3

`4/15+35/99+11/15-2/7+64/99-5/7`

`=(4/15+11/15)+(35/99+64/99)-(2/7+5/7)`

`=15/15+99/99-7/7`

`=1+1-1`

`=2-1`

`=1`

1

Ta có: \(\frac{4}{15}+\frac{35}{99}+\frac{11}{15}-\frac27+\frac{64}{99}-\frac57\)

\(=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{35}{99}+\frac{64}{99}\right)-\left(\frac27+\frac57\right)\)

\(=\frac{15}{15}+\frac{99}{99}-\frac77\)

=1+1-1

=1

\(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+\frac{4}{11}\cdot\frac{4}{15}+...+\frac{4}{95}\cdot\frac{4}{99}\)

\(A=\frac{16}{3\cdot7}+\frac{16}{7\cdot11}+\frac{16}{11\cdot15}+...+\frac{16}{95\cdot99}\)

\(A=4\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{95\cdot99}\right)\)

\(A=4\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=4\cdot\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=4\cdot\frac{32}{99}\)

\(A=\frac{128}{99}\)

\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{95}\times\frac{4}{99}\)

     \(=4\times\frac{4}{3.7}+4\times\frac{4}{7.11}+...+4\times\frac{4}{95.99}\)

     \(=4\times\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)

     \(=4\times\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)

     \(=4\times\left(\frac{1}{3}-\frac{1}{99}\right)\)

     \(=4\times\frac{32}{99}\)

     \(=\frac{128}{99}\)