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6 tháng 5 2016
Gọi tổng đó là tổng S
Ta có: S = 1/6+1/12+1/30+1/42+1/56+1/72+1/90
=> S = 1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=> S = 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
=> S = 1/2-1/10
=> S = 5/10-1/10
=> S=4/10
=> S=2/5
24 tháng 7 2019
\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)
\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)
\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)
\(\Rightarrow B=3\cdot\frac{98}{99}\)
\(\Rightarrow B=\frac{98}{33}\)
24 tháng 7 2019
\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)
\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow A=1-\frac{1}{7}\)
\(\Rightarrow A=\frac{6}{7}\)
PT
2
1 tháng 5 2016
\(S1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(S1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(S1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(S1=1-\frac{1}{10}\)
\(S1=\frac{9}{10}\)
CHÚC BN HC GIỎI !!!!!!!!!! TỨ DIỆP THẢO
1 tháng 5 2016
S=\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{9.10}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...............+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
NM
1
MT
4 tháng 6 2020
Bổ sung đề bài : Tính
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
VN
30 tháng 1 2023
\(1)\)
\(a)(-50).37=-1850\)
\(b)(-20).(-14).25=280.25=7000\)
\(c)40.(-30).12=-1200.12=-14400\)
\(2)\)
\(a)\)
\(-40.[(-15)+30)]\)
\(=-40.15\)
\(=-600\)
\(b)\)
\(-5 . [(-27) - (-7)]\)
\(=-5 . [(-27)+7]\)
\(=-5 . (-20)\)
\(=100\)
\(c)\)
\(-10 . [(-28) + (-22) - 10]\)
\(=-10 . [-50 - 10]\)
\(=-10.(-60)\)
\(=600\)
16 tháng 3 2018
Ta có: \(B=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(=\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+...+\frac{11-10}{10\cdot11}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{11-3}{3\cdot11}=\frac{8}{33}\)
Vậy \(B=\frac{8}{33}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2550}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2550}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{50\cdot51}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)