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LC
3
8 tháng 8 2020
Bài làm:
Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)
\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)
\(=\frac{1}{2}.\frac{11}{12}\)
\(=\frac{11}{24}\)
NH
8 tháng 8 2020
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=2\times\left(1-\frac{1}{12}\right)\)
\(=2\times\frac{11}{12}\)
\(=\frac{11}{6}\)
13 tháng 7 2023
=2(1/12+1/30+...+1/132)
=2(1/3-1/4+1/5-1/6+1/6-1/7+...+1/11-1/12)
=2(1/12+1/5-1/12)
=2*1/5=2/5
BN
1
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)
\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)
\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)
\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)
\(A=2\times\dfrac{9}{22}\)
\(A=\dfrac{9}{11}\)
VB
1
NH
5
8 tháng 6 2018
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{1}{5}=\frac{1}{4}\)
8 tháng 6 2018
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}\)
\(\Rightarrow A=\frac{2}{5}.2\)
\(\Rightarrow A=\frac{4}{5}\)
TL
2 tháng 12 2016
B1 Nhân biểu tthức trên với 1/2 được
1/6+1/12/+1/20+....+1/110=1/2×3+1/3×4+.....+1/10×11
=1/2-1/3+1/3-1/4+.....+1/10-1/11=1/2-1/11=9/22
B2
B.1/1×2+1/2×3+.....+1/99×100=1-1/2 +1/2-1/3+....+1/99-1/100=1-1/100=99/100
Phần a sai đề phải llà1/2×4+1/4×6+.....+1/8×10 mới làm đc nhé