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\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)
Vậy ................
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{9999}{10000}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{99.101}{100.100}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right).\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right).\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(B=\frac{1\cdot2\cdot3\cdot..\cdot99}{2\cdot3\cdot4\cdot..\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(B=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)
vậy......
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/3.(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17+3/17.20)
A=1/3.(1/2-1/20)
=3/20
B=1.3/2.2+2.4/3.3+3.5/4.4+...+99.101/100.100
B=(1.2.3...99).(3.4.5...101)/(2.3.4...100).(2.3.4...100)
B=\(\frac{1.2....99}{2.3...100}\).\(\frac{3.4...101}{2.3...100}\)
B=1/100.101/2=101/200
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\frac{9}{20}\)
\(=\frac{3}{20}\)
\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
a, \(A=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{340}\)
\(\Leftrightarrow A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{17.20}\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{17.20}\right)\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{6}-\frac{1}{60}=\frac{3}{20}\)
b, \(2004^{10}+2004^9=2004^9\left(2014+1\right)=2014^9+2005\)
\(2015^{10}=2015^9.2015\)
-Vậy: \(2004^{10}+2004^9< 2005^{10}\)
\(\frac{1}{10}\) + \(\frac{1}{40}\) +\(\frac{1}{88}\)+\(\frac{1}{154}\) + \(\frac{1}{238}\) + \(\frac{1}{340}\)
= \(\frac{1}{8}\) + \(\frac{1}{56}\) + \(\frac{1}{140}\)
= \(\frac{1}{7}\) + \(\frac{1}{140}\)
= \(\frac{3}{20}\)
S=1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 11/17.20
S=1/3.(1/2-1/5+1/5-1/8+.......+1/17 - 1/20)
S=1/3.(1/2-1/20)
S=1/3.9/20
S=3/20
Ai tích mk mk sẽ tích lại
Chú ý tích gấp ddooi khi tích đủ 3 cái
\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)
\(suyra:A<1\)