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21 tháng 3 2019
cậu tham khảo ở đây
https://olm.vn/hoi-dap/detail/204681226462.html
6 tháng 7 2019
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
BK
6 tháng 7 2019
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
XO
7 tháng 7 2019
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\frac{98}{99}=0\)
\(\Rightarrow3x=0+\frac{98}{99}\)
\(\Rightarrow3x=\frac{98}{99}\)
\(\Rightarrow x=\frac{98}{99}:3\)
\(\Rightarrow x=\frac{98}{297}\)
7 tháng 7 2019
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)
\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)
\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)
\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(6x-\left(1-\frac{1}{11}\right)=0\)
\(6x-\frac{10}{11}=0\)
\(6x=\frac{10}{11}\)
\(x=\frac{5}{33}\)
26 tháng 6 2016
\(A=1+1+1+1-\left(\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}\right)\)
\(A=4+\left(\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}\right)\)
\(A=4+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(A=4+\left(1-\frac{1}{9}\right)\)
\(A=4+\frac{8}{9}=\frac{44}{9}\)
Vậy A=44/9
M
4 tháng 10 2017
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right).\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{171}{595}.\frac{-4}{3}}\)
\(M=\frac{-1}{12}:\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
D
9 tháng 9 2018
Ta có:
\(M=\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{1}{17}+\frac{1}{7}-\frac{-3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\left(\frac{1}{30}-\frac{7}{20}\right)\times\frac{5}{19}}{\left(\frac{24}{119}+\frac{3}{35}\right)\times\frac{-4}{3}}\)
\(M=\frac{\frac{-19}{60}\times\frac{5}{19}}{\frac{171}{595}\times\frac{-4}{3}}\)
\(M=\frac{-1}{12}\div\frac{-228}{595}\)
\(M=\frac{595}{2736}\)
Vậy \(M=\frac{595}{2736}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 5 2022
Sửa đề: \(M=\dfrac{\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right)\cdot\dfrac{5}{19}}{\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{-3}{35}\right)\cdot\dfrac{-4}{45}}\)
\(=\dfrac{\dfrac{3\cdot6-4\cdot4-7\cdot3}{60}\cdot\dfrac{5}{19}}{\dfrac{7+5+3}{35}\cdot\dfrac{-4}{45}}=\dfrac{\dfrac{-19}{60}\cdot\dfrac{5}{19}}{\dfrac{15}{35}\cdot\dfrac{-4}{45}}=\dfrac{-1}{12}:\dfrac{-4}{105}=\dfrac{105}{60}=\dfrac{7}{4}\)
5 tháng 1 2016
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99\cdot101}\right)\)
\(2A=-\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{99}-\frac{2}{101}\right)\)
\(2A=-\left(2-\frac{2}{101}\right)\)
\(2A=-\frac{200}{101}\)
\(\Rightarrow A=-\frac{100}{101}\)
28 tháng 2 2018
Đặt biểu thức trên là A, ta có:
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(\Rightarrow A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow A=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\frac{100}{101}\)
\(\Rightarrow A=-\frac{100}{101}\div2=-\frac{50}{101}\)
Đặt \(A=-\frac13+\left(-\frac{1}{15}\right)+\left(-\frac{1}{35}\right)+\left(-\frac{1}{63}\right)+\cdots+\left(-\frac{1}{999999}\right)\)
\(-A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\cdots+\frac{1}{999\cdot1001}\)
\(-A=\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{999}-\frac{1}{1001}\)
\(-A=\frac11-\frac{1}{1001}\)
\(-A=\frac{1000}{1001}\)
\(\Rightarrow A=-\frac{1000}{1001}\)
Vậy \(A=-\frac{1000}{1001}\)