Lời giải:

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

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

$=99.100.101-9.10.11$

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

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!

ta có : 1.2+2.3+3.4+.....+99.100=99.100.101 /3 =333300

mà 1.2+2.3+....+9.10+9.10.11/3=330

=>E= 333300-330=332970

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

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

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

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

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

\(x\)    = 11

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

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

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

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

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

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

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

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

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

=1/10-1/100

=9/100