1/11.12.13 + 1/12.13.14 +...+1/30.31.32
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Những câu hỏi liên quan
9 tháng 5 2016
Đặt A=\(\frac{-1}{2\cdot3\cdot4}\)+\(\frac{-1}{3\cdot4\cdot5}\)+...+\(\frac{-1}{12\cdot13\cdot14}\)
A=(-1)*(\(\frac{1}{2\cdot3\cdot4}\)+\(\frac{1}{3\cdot4\cdot5}\)+...+\(\frac{1}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{12\cdot13}\)-\(\frac{1}{13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{6}\)-\(\frac{1}{182}\))
A=\(\frac{-1}{2}\)*\(\frac{44}{273}\)
A=\(\frac{-22}{273}\)
9 tháng 5 2016
\(\frac{-1}{2.3.4}+\frac{-1}{3.4.5}+\frac{-1}{4.5.6}+...+\frac{-1}{11.12.13}+\frac{-1}{12.13.14}\)
\(=-\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{11.12.13}+\frac{2}{12.13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{11.12}-\frac{1}{12.13}+\frac{1}{12.13}-\frac{1}{13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{13.14}\right)=-\frac{1}{2}.\frac{44}{273}=-\frac{22}{273}\)
2 tháng 7 2016
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.50}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
HP
2 tháng 7 2016
5,có thể bn sai đề chỗ \(\frac{1920}{111.2013}\),phải là \(\frac{1902}{111.2013}\)
\(F=\frac{3}{15.18}+\frac{11}{18.29}+\frac{12}{29.41}+\frac{70}{41.111}+\frac{1902}{111.2013}\)
\(=\frac{18-15}{15.18}+\frac{29-18}{18.29}+\frac{41-29}{12}+\frac{111-41}{41.111}+\frac{2013-111}{1902}\)
\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}+\frac{1}{41}-\frac{1}{111}+\frac{1}{111}-\frac{1}{2013}\)
\(=\frac{1}{15}-\frac{1}{2013}=\frac{222}{3355}\)
PN
0
22 tháng 3 2017
Ta có công thức:
\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)
vậy
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)
mình làm đầu tiên đó,
Chúc bạn học tốt !
S
22 tháng 3 2017
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)
\(=\frac{1}{2}\cdot\frac{77}{156}\)
\(=\frac{77}{312}\)
ZH
1
18 tháng 9 2015
Câu 1:
Đặt S = 1.2+2.3+3.4+...+30.31
3 S = 1.2.3+2.3.3+3.4.3+...+30.31.3
3 S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ...+ 30.31.(32-29)
3S = 1.2.3 + 2.3.4-2.3 + 3.4.5-2.3.4 + ...+ 30.31.32-29.30.31
3S= 30.31.32
S= 30.31.32/3
DP
1 tháng 8 2017
\(\frac{4}{7}x+\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{12.13.14}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{12.13}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\frac{45}{91}\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\frac{45}{182}=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{39}{40}-\frac{45}{182}\Leftrightarrow\frac{4}{7}x=\frac{2649}{3640}\)
\(\Rightarrow x=\frac{2649}{3640}\div\frac{4}{7}=\frac{2649}{2080}\)
Vậy x = \(\frac{2649}{2080}\)