Tìm x biết
5/1.7+5/7.13+5/13.19+...+5/x.(x+6)=10075/12096
K
Khách
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13 tháng 8 2018
Ta có :\(\left(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{601.607}\right)\)\(\ne0\)
\(\Rightarrow x=0\)
13 tháng 8 2018
\(X:\left(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+......+\frac{5}{601.607}\right)=0\)
\(\Rightarrow X:\left(\frac{5}{1}-\frac{5}{7}+\frac{5}{7}-\frac{5}{13}+\frac{5}{13}+......+\frac{5}{601}-\frac{5}{607}\right)=0\)
\(\Leftrightarrow X:\left(5-\frac{5}{607}\right)=0\)
\(\Leftrightarrow X:\frac{3030}{607}=0\)
\(\Leftrightarrow X=0\)
CÁCH 2:\(X:\left(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+....+\frac{5}{601.607}\right)=0\)
\(\Leftrightarrow X=0.\left(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+....+\frac{5}{601.607}\right)\)
\(\Leftrightarrow X=0\)
HT
1 tháng 7 2016
\(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{2017.2023}\)
\(=5.\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{2017.2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{2017}-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\frac{2022}{2023}\)
\(=\frac{1685}{2023}\)
10 tháng 7 2016
\(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{2017.2023}\)
\(=\frac{5.6}{1.7.6}+\frac{5.6}{7.13.6}+\frac{5.6}{13.19.6}+.....+\frac{5.6}{2017.2023.6}\)
\(=\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{2017.2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{2017}-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\frac{2022}{2023}\)
\(=\frac{1685}{2023}\)
22 tháng 7 2017
E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)
\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)
\(=6\cdot\frac{99}{100}=\frac{297}{50}\)
F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)
G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
CN
1
16 tháng 2 2017
ta có:
\(S=\frac{4}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{43.49}\right)\)\(=\frac{4}{6}\left(\frac{7-1}{1.7}+\frac{13-7}{7.13}+...+\frac{49-43}{43.49}\right)=\frac{4}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{43}-\frac{1}{49}\right)\)
\(\frac{4}{6}\left(1-\frac{1}{49}\right)=\frac{4.48}{6.49}=\frac{32}{49}\)
LT
1
25 tháng 6 2015
1/7+1/91+1/247+1/475+1/775+1/1147=? (1)
ta có: (1) <=>: 1/(1.7)+1/(7.13)+1/(13.19)+1/(19.25)+1/(25.31)+1/(31.37)
=1/6.(1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
=1/6.(1-1/37)=6/37
DV
1
WR
15 tháng 6 2015
G=6(6/1.7+6/7.13+6/13.19+..+6/n(n+6) )
=6(1-1/7+1/7-1/13+1/13-1/19+....+1/n-1/n+6)
=6(1-n/n+6)
=6.6/n+6
=36/n+6
vậy G=36/n+6
S
4 tháng 5 2023
a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)
Đặt \(A=\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+....+\frac{5}{x.\left(x+6\right)}\)
\(\Rightarrow A=\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{x.\left(x+6\right)}\right)\)
\(\Rightarrow A=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{x}-\frac{1}{x+6}\right)\)
\(\Rightarrow A=\frac{5}{6}.\left(1-\frac{1}{x+6}\right)\)
\(\Rightarrow\frac{5}{6}.\frac{x+5}{x+6}=\frac{10075}{12096}\)
Làm nốt nha
\(\frac{5}{1.7}+\frac{5}{7.13}+...+\frac{5}{x.\left(x+6\right)}=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{x.\left(x+6\right)}\right)=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+6}\right)=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{x+6}\right)=\frac{10075}{12096}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}:\frac{5}{6}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}.\frac{6}{5}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{x+6}=1-\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{x+6}=\frac{1}{2016}\)
\(\Rightarrow x+6=2016\)
\(\Rightarrow x=2016-6\)
\(\Rightarrow x=2010\)
Chúc bạn học tốt !!!