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NT
3
23 tháng 8 2016
=4x(\(\frac{1}{11x13}\)+\(\frac{1}{13x15}\)+.......+\(\frac{1}{99x101}\))
=4x(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\))
4x(\(\frac{1}{11}\)-\(\frac{1}{101}\))
=4x \(\frac{90}{1111}\)
=\(\frac{360}{1111}\)
23 tháng 8 2016
\(\frac{4}{11\times13}+\frac{4}{13\times15}+\frac{4}{15\times17}+...+\frac{4}{99\times101}\)
\(=\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+\frac{4}{15}-\frac{4}{17}+...+\frac{4}{99}-\frac{4}{101}\)
\(=\frac{4}{11}-\frac{4}{101}\)
\(=\frac{360}{1111}\)
DP
6 tháng 8 2017
\(\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19}+\frac{2}{19.21}\right)\cdot462-x=19\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)\cdot462-x=19\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{21}\right)\cdot462-x=19\)
\(\Rightarrow\frac{10}{231}\cdot462-x=19\)
\(\Rightarrow20-x=19\Rightarrow x=1\)
TP
6 tháng 8 2017
Ta có:
\(\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19}+\frac{2}{19.21}\right).462-x=19\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right).462-x=19\)
\(\Rightarrow\left(\frac{1}{11}-\frac{1}{21}\right).462-x=19\)
\(\Rightarrow\frac{10}{231}.462-x=19\Leftrightarrow20-x=19\)
\(\Rightarrow x=20-19=1\)
4 tháng 7 2018
\(\frac{3}{11\text{x}13}+\frac{3}{13\text{x}15}+\frac{3}{15\text{x}17}+\frac{3}{17\text{x}19}+\frac{3}{19\text{x}21}\)
\(=\frac{3}{2}\text{x}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{3}{2}\text{x}\left(\frac{1}{11}-\frac{1}{21}\right)\)
\(=\frac{3}{2}\text{x}\frac{10}{231}\)
\(=\frac{5}{77}\)
H
0
H
0
NN
2
S
18 tháng 5 2017
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Rightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
Đặt S = \(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+....+\frac{20}{53.55}\)
\(=10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)
\(=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=10\left(\frac{1}{11}-\frac{1}{55}\right)\)
\(=10\cdot\frac{4}{55}=\frac{8}{11}\)
Thay A vào đề bài,ta được:
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{8}{11}\)
\(x=1\)
\(A=13.15+15.17+17.19+...+99.101\)
\(\Rightarrow6A=13.15.6+15.17.6+17.19.6+...+99.101.6\)
\(\Rightarrow6A=13.15.\left(17-11\right)+15.17.\left(19-13\right)+17.19.\left(21-15\right)+...+99.101.\left(103-97\right)\)
\(\Rightarrow6A=\left(13.15.17-11.13.15\right)+\left(15.17.19-13.15.17\right)+\left(17.19.21-15.17.19\right)+...+\left(99.101.103-97.99.101\right)\)
\(\Rightarrow6A=99.101.103-11.13.15\)
\(\Rightarrow6A=1027752\)
\(\Rightarrow A=171292\)