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Tính giá trị biểu thức :
C = \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
K
Khách
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Những câu hỏi liên quan
24 tháng 10 2016
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
HP
2
S
9 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\)
2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}\)
2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)
2A = \(1-\frac{1}{9}=\frac{8}{9}\)
A = \(\frac{8}{9}:2=\frac{4}{9}\)
Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\)
2B = \(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\)
2B = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
2B = \(\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
B = \(\frac{2}{5}:2=\frac{1}{5}\)
Thay A và B vào S ta được:
\(S=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
Q
9 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
\(\Rightarrow S=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)\)
\(S=\frac{1}{2}.\frac{58}{45}\)
\(S=\frac{29}{45}\)
8 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
MA
7 tháng 9 2016
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
HT
7 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
9 tháng 5 2017
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+.....+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{8}{9}+\frac{1}{2}\cdot\frac{2}{5}=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
9 tháng 5 2017
=1/2x(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.....+1/8-1/10)
=1/2x58/45
=29/45
TP
8 tháng 8 2017
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
R
24 tháng 8 2019
Bài làm
D=ko viết lại đề
=1/1.3+1/1.5+1/5.7+1/7.9-1/2.4-1/4.6-1/6.8-1/8.10
=1+1/9-1-1/10
=10/9-9/10
=19/90
24 tháng 8 2019
=(1/1.3+...+1/7.9)-(1/2.4+...+1/8.10)
=2(1/1.3+...+1/7.9)-2(1/2.4+...+1/8.10)
=(2/1.3+...+2/7.9)-(2/2.4+...+2/8.10)
=(1-1/3+...+1/7-1/9)-(1/2-1/4+ +1/8-1/10)
=1-1/9-1/2+1/10
tự tính tiếp nhé
TM
2
LH
28 tháng 2 2018
2c=2/1.3+2/2.4+2/3.5+...+2/2016.2018
2c=1-1/3+1/2-1/4+1/3-1/5+...+1/2016-1/2017
2c=1-1/2017
2c=2016/2017
c=4032/2017
11 tháng 4 2019
Lê Minh Hiếu cách lm đúng mà sai kết quả
KQ : \(\frac{2016}{4034}\)