\(A=\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{9901}{9900}=\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+\left(1+\frac{1}{4.5}\right)+...+\left(1+\frac{1}{99.100}\right)\)\(=\left(1+1+1+...+1\right)+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=98+\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=98+\frac{49}{100}=98\frac{49}{100}\)

Mấy câu như này tách ra kiểu gì?

\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)

\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{71}{72}+\frac{89}{90}=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=8-\frac{2}{5}=\frac{38}{5}\)

A=1+2+3+4+5+...+99+100

A=(1+100).100:2=101.50=5050

B=1/2+1/6+1/12+1/20+1/30+...+1/9900

B=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100

B=1-1/100=99/100

A = 100 x 101 : 2 = 5050

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)

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

      \(=1-\frac{1}{100}\)

        \(=\frac{99}{100}\)

99/100 

có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi

\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)

\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)

Trả lời:

2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(K=\frac{1}{2}-\frac{1}{100}\)

\(K=\frac{49}{100}\)

3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)

\(L=\frac{5}{3}\times\frac{102}{103}\)

\(L=\frac{170}{103}\)

Học tốt 

2)\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

3)\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)