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6 tháng 3 2023
\(=\dfrac{1}{7}+\dfrac{4}{6}\left(\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{1}{7}+\dfrac{2}{3}\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{7}+\dfrac{2}{3}\cdot\dfrac{6}{49}=\dfrac{1}{7}+\dfrac{4}{49}=\dfrac{11}{49}\)
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
\(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{2017.2023}\)
\(=\frac{5.6}{1.7.6}+\frac{5.6}{7.13.6}+\frac{5.6}{13.19.6}+.....+\frac{5.6}{2017.2023.6}\)
\(=\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{2017.2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{2017}-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\left(1-\frac{1}{2023}\right)\)
\(=\frac{5}{6}.\frac{2022}{2023}\)
\(=\frac{1685}{2023}\)
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
NN
1 tháng 4 2019
\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+...+\frac{1}{61.67}\)
=6.\(\left(\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{61.67}\right)\):6
=\((\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{61.67}):6\)
=\(\left(1-\frac{1}{7}+\frac{1}{7}+\frac{1}{13}+...+\frac{1}{61}+\frac{1}{67}\right):6\)
=\(\left(1-\frac{1}{67}\right):6\)
=\(\frac{66}{67}:6=\frac{66}{67}.\frac{1}{6}=\frac{11}{67}\)
NN
3
10 tháng 4 2018
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
AK
10 tháng 4 2018
\(\frac{5}{1.7}+\frac{5}{7.13}+...+\frac{5}{x.\left(x+6\right)}=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{x.\left(x+6\right)}\right)=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+6}\right)=\frac{10075}{12096}\)
\(\Rightarrow\frac{5}{6}.\left(1-\frac{1}{x+6}\right)=\frac{10075}{12096}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}:\frac{5}{6}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{10075}{12096}.\frac{6}{5}\)
\(\Rightarrow1-\frac{1}{x+6}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{x+6}=1-\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{x+6}=\frac{1}{2016}\)
\(\Rightarrow x+6=2016\)
\(\Rightarrow x=2016-6\)
\(\Rightarrow x=2010\)
Chúc bạn học tốt !!!
ta có:
\(S=\frac{4}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{43.49}\right)\)\(=\frac{4}{6}\left(\frac{7-1}{1.7}+\frac{13-7}{7.13}+...+\frac{49-43}{43.49}\right)=\frac{4}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{43}-\frac{1}{49}\right)\)
\(\frac{4}{6}\left(1-\frac{1}{49}\right)=\frac{4.48}{6.49}=\frac{32}{49}\)