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= 5.(1/2 + 1/6+1/12+1/20+...+1/9506+1/9702+1/9900)

 = 5. (1/1.2 + 1/2.3+1/3.4+1/4.5+...1/97.98+1/98.99+1/99.100)

= 5 .(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/97-1/98+1/98-1/99+1/99-1/100)

= 5.(1-1/100)=5. 99/100=99/20

25

5144 nhe

Hơi nhầm nè , để tôi sửa lại đề \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\)

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

\(A=1+1+1+...+1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-....-\frac{1}{9900}\)

\(A=98-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\right)\)

\(A=98-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

\(A=98-\left(\frac{1}{2}-\frac{1}{100}\right)=98-\frac{49}{100}=\frac{9751}{100}\)

Vậy.............

 \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9989}{9900}\)

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

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

               có 50 số 1

\(A=50-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

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

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

Thay B vào A ta được:

\(A=50-\frac{49}{100}=\frac{5000}{100}-\frac{49}{100}=\frac{4951}{100}\)

cảm ơn 2 bạn nhiều nha !

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 

Ta có : 

\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)

\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)

\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)

\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)

\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)

\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)

\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)

Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra : 

\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\frac{A}{100}=98-\frac{49}{100}\)

\(\frac{A}{100}=\frac{9751}{100}\)

\(A=\frac{9751}{100}.100\)

\(A=9751\)

Vậy \(A=9751\)

Chúc bạn học tốt ~ 

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

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

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

\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+...+\frac{5}{90}\)( viết tắt )

\(I=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{9.10}\)

\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

\(I=5\times\frac{2}{5}\)

\(I=2\)

Vậy \(I=2\)

Tk nha bn ~~

\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)

\(I=\frac{5}{2\cdot3}+\frac{5}{3\cdot4}+\frac{5}{4\cdot5}+\frac{5}{5\cdot6}+\frac{5}{6\cdot7}+\frac{5}{7\cdot8}+\frac{5}{8\cdot9}+\frac{5}{9\cdot10}\)

\(I=5\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)

Theo tính chất của toán HSG lớp 6, ta được

\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(I=5\left(\frac{5}{10}-\frac{1}{10}\right)\)

\(I=5\cdot\frac{4}{10}=5\cdot\frac{2}{5}=\frac{10}{5}=2\)

\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)

\(I=5\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(I=5.\frac{2}{5}\)

\(I=2\)