\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{1}{4850}\)
K
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EL
2
17 tháng 8 2016
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{1}{4850}\)
\(=\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+...+\frac{2}{9700}\)
\(=2.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
CK
6
7 tháng 3 2016
Ta có :
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
Giá trị không đổi khi cả tử và mẫu cùng nhân với 2, ta được :
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
7 tháng 3 2016
\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{1}{2}.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{19}\right)=\frac{9}{19}\)
KG
6
7 tháng 3 2017
N = 2/4+2/28+2/70+2/130+2/208+2/304
N = 2/1.4+2/4.7+2/7.10+2.......
DM
6
VT
C=\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
trình bày mới tk
4
12 tháng 7 2017
\(C=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(C=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(C=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
\(C=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)
\(C=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
BD
7 tháng 3 2017
M=1/2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19
M=1/2-1/2+1/7-1/7+1/5-1/5+1/13-1/13+1/8-1/8+1/19
M=1/2-1/19
M=17/38
B
1
CK
2
19 tháng 9 2020
Ta có: \(\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{2}{x^2+3x}=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}\cdot\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{2}{3}\cdot\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{2}{3}\cdot\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{x+3}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{x+3}=\frac{1}{9}:\frac{2}{3}=\frac{1}{9}\cdot\frac{3}{2}=\frac{1}{6}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{6}-\frac{1}{4}=\frac{2}{12}-\frac{3}{12}=-\frac{1}{12}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{12}\)
\(\Leftrightarrow x+3=12\)
hay x=9
Vậy: x=9
Đặt tổng A ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{35}+...+\frac{1}{4850}\)
\(\frac{3}{2}A=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{1}{130}+...+\frac{3}{9700}\)
\(\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)
\(\frac{3}{2}A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{33}{50}\)