tính nhanh : 9.10 + 10.11 + ... + 17.18 + 18.19
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Barack mới tự học toán 6 hơn 1 tuần, có cách giải này
= 102 + 10 + 112 + 11 + 122 + 12 + 132 + 13 + 142 + 14 + 152 + 15 +162 +16 +172 +17 + 182 + 18 + 192 +19
= 102 + 112 + 132 + 142 + 152 + 162 + 172 + 182 +192 + 10 + 11 +12 +13 + 14 + 15 + 16 +17 + 18 + 19
5/15.16+5/16.17+5/17.18+5/18.19+5/19.20
=5(1/15-1/16+1/16-1/17+..+1/19-1/20)
=5.(1/15-1/20)
=5.1/60
=1/12
\(\frac{5}{15.16}+\frac{5}{16.17}+\frac{5}{17.18}+\frac{5}{18.19}+\frac{5}{19.20}\)
= \(5.\left(\frac{1}{15.16}+\frac{1}{16.17}+\frac{1}{17.18}+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
= \(5.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+\frac{1}{17}-\frac{1}{18}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
= \(5.\left(\frac{1}{15}-\frac{1}{20}\right)\)
= \(5.\frac{1}{60}\)
= \(\frac{1}{12}\)
\(M=\dfrac{15}{7\cdot8}-\dfrac{17}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(M=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(M=\dfrac{1}{7}-\dfrac{1}{11}=\dfrac{11-7}{7\cdot11}=\dfrac{4}{77}\)
`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`
\(\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}-\frac{19}{9.10}+\frac{21}{10.11}\)
\(=\frac{3+4}{3.4}-\frac{4+5}{4.5}+\frac{5+6}{5.6}-\frac{6+7}{6.7}+\frac{7+8}{7.8}-\frac{8+9}{8.9}-\frac{9+10}{9.10}+\frac{10+11}{10.11}\)
\(=\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{8}-\frac{1}{9}+\frac{1}{9}+\frac{1}{10}-\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
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 . ( \(\frac{1}{9}-\frac{1}{x+1}\)) = \(\frac{2}{9}-\frac{2}{x+1}\)