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Đặt : \(A=10.11+11.12+...+98.99+99.100\)

\(\Rightarrow3A=10.11.3+11.12.3+...+98.99.3+99.100.3\)

\(\Rightarrow3A=10.11.\left(12-9\right)+11.12.\left(13-10\right)+...+\)\(98.99.\left(100-97\right)+99.100.\left(101-98\right)\)

Lời giải:

$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$

$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$

$=99.100.101-9.10.11$

$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$

giúp tớ với

Số số hạng là(99,100-10,11):0,01+1=8900

=>Tổng=8900:2x(10,11+99,100)=485984,5

Ta có A=1/10.11+1/11.12+...+1/98.99+1/99.100

            =1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100

            =1/10-1/100

           =10/100-1/100

          =9/100

Vậy A=9/100

Giải:

A=1/10.11+1/11.12+...+1/98.99+1/99.100

A=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100

A=1/10-1/100

A=9/100

Chúc bạn học tốt!

10.11+11.12+12.13+...+97.98+98.99+99.100

=10-11+11-12+12-13+...+97-98+98-99+99-100

=10-100

=-90

Đặt A = 10.11 + 11.12 + ... + 98.99 + 99.100 

3A = 10.11.3 + 11.12.3 + ... + 98.99.3 + 99.100.3 

3A = 10.11.(12 -9) + 11.12.(13-10) + ... + 98.99.(100 - 97) + 99.100.(101-98)

3A = 10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + ... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100 

3A = (10.11.12 + 11.12.13 + ... + 98.99.100 + 99.100.101) - (9.10.11 + 10.11.12 + ... + 97.98.99 + 98.99.100)

3A = 99.100.101 - 9.10.11 

3A = 999799 

A = 999799 : 3  

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!

\(\dfrac{x}{10.11}\) + \(\dfrac{x}{11.12}\) +................+ \(\dfrac{x}{99.100}\)= \(\dfrac{99}{100}\)

\(x\)( \(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\) +..........+\(\dfrac{1}{99.100}\)) = \(\dfrac{99}{100}\)

\(x\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{13}\)+...........+\(\dfrac{1}{99}\)- \(\dfrac{1}{100}\)) = \(\dfrac{99}{100}\)

\(x\). \(\dfrac{9}{100}\)           = \(\dfrac{99}{100}\)

\(x\)     = \(\dfrac{99}{100}\) : \(\dfrac{9}{100}\)

\(x\)    = 11