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7 tháng 8 2016
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{6}-\frac{1}{15}\)
\(A=\frac{1}{10}\)
7 tháng 8 2016
A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)
=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)
20 tháng 7 2016
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)
\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)
20 tháng 7 2016
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{6}-\frac{1}{15}\)
\(A=\frac{1}{10}\)
NV
2
KV
4 tháng 10 2019
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(\Rightarrow S=1-\frac{1}{46}\)
\(\Rightarrow S=\frac{45}{46}\)
4 tháng 10 2019
Bài làm
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(S=\frac{1}{1}-\frac{1}{46}\)
\(S=\frac{46}{46}-\frac{1}{46}\)
\(S=\frac{45}{46}\)
Vậy \(S=\frac{45}{46}\)
# Học tốt #
BM
10 tháng 4 2019
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
13 tháng 8 2015
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{12.13}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{4}-\frac{1}{13}=\frac{9}{52}\)
S
10 tháng 5 2017
Bài 1:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
Bài 2:
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vậy A < 2
Bài 3:
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Bài 4:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)
10 tháng 5 2017
A=1-1/2+1/2-1/3+.............................1/49-1/50
A=1-1/50
A=49/50
NN
2
5 tháng 5 2016
A=1/2+1/6+1/12+...+1/132
=1/1.2+1/2.3+1/3.4+...+1/11.12
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/11-1/12
=1-1/12
=11/12
5 tháng 5 2016
A=1/2+1/6+1/12+...+1/132
=1/1.2+1/2.3+1/3.4+...+1/11.12
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/11-1/12
=1-1/12
=11/12
Vậy A=11/12
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}+\frac{1}{156}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}+\frac{1}{12.13}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)
\(S=\frac{1}{1}-\frac{1}{13}\)
\(S=\frac{12}{13}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)