\(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}\)

\(=\frac{8-7}{7\cdot8}+\frac{9-8}{8\cdot9}+\frac{10-9}{9\cdot10}+\frac{11-10}{10\cdot11}+\frac{12-11}{11\cdot12}+\frac{13-12}{12\cdot13}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)

\(=\frac{1}{7}-\frac{1}{13}=\frac{13-7}{7\cdot13}=\frac{6}{91}\)

\(\frac{6}{91}\)nha

\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)

\(\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}.\frac{1}{2}\)

\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)

\(\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}=\frac{1}{9}\)

=>x+1=9

=>x=8

\(M=\dfrac{15}{7\cdot8}-\dfrac{17}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)

\(M=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)

\(M=\dfrac{1}{7}-\dfrac{1}{11}=\dfrac{11-7}{7\cdot11}=\dfrac{4}{77}\)

`M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`

`=1/7-1/8-(1/8-1/9)+1/9-1/10+1/10-1/11`

`=1/7-1/4+2/9-1/11`

`=67/2772`

Theo mình đề bài sai phải là `M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`

`=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11`

`=1/7-1/11=4/77`

M=18+17−18−19+19−110+110−111M=18+17−18−19+19−110+110−111

ko ghi lại đề

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)

\(=1-\frac{1}{212}\)

\(=\frac{211}{212}\)

\(\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{211.212}\)

= \(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{211}-\frac{1}{212}\)

= \(\frac{1}{8}-\frac{1}{212}\)

= \(\frac{51}{424}\)

\(\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{210.211}+\frac{1}{211.212}\)

=\(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)

=\(\frac{1}{8}-\frac{1}{212}\)=\(\frac{53}{424}-\frac{2}{424}=\frac{51}{424}\)\(\approx0,12028\approx0,12\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{2}-\frac{1}{15}\)

\(=\frac{13}{30}\)

co \(\frac{1}{9\cdot10}=\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)

............

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

nen \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\)

\(=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-...+\frac{1}{x}-\frac{1}{x+1}\)

=\(\frac{1}{9}-\frac{1}{x+1}\)

2 .  ( \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\))

=  2  . ( \(\frac{1}{9}-\frac{1}{x+1}\))  = \(\frac{2}{9}-\frac{2}{x+1}\)

Bạn ơi tim x mà bạn

bo may ko biet

1003002000

cách giải ?

Câu hỏi của Phung Ngoc Quoc Bao - Toán lớp 6 - Học toán với OnlineMath

Cách thực hiện y hệt 

giai dum voi dang nay minh lam lan dau tien len khong biet

Nhan B voi 3 ta duoc:

3B=9.10.3+10.11.3+11.12.3+.....+98.99.3

3B=9.10.(11-8)+10.11.(12-9)+11.12.(13-3)+.....+98.99.(100-97)

3B=(9.10.11-8.9.10)+(10.11.12-9.10.11)+(11.12.13-10.11.12)+.....+(98.99.100-97.98.99)

3B=98.99.100-8.9.10=970200-720=969480

B=969480:3=323160

ok roi ban nhe!