\(\frac{2}{4\cdot7}-\frac{2}{5\cdot9}+\frac{2}{7\cdot10}-\frac{2}{9\cdot13}+\frac{2}{10\cdot13}-\frac{2}{13\cdot17}+...+\frac{2}{301\cdot304}-\frac{2}{401\cdot405}CM:tich,tren,< \frac{1}{15}\)
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\(C=\frac{2}{1\cdot5}+\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+\frac{2}{17\cdot21}\)
\(C=\frac{2}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
ta có: \(B=\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(=\frac{1}{2}.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{401.405}\right)\)
\(=\frac{1}{2}.\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{401.405}\right)\)
\(=\frac{1}{8}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{17}-...-\frac{1}{401}+\frac{1}{405}\right)\)
\(=\frac{1}{8}.\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(=\frac{1}{8}.\frac{80}{405}=\frac{10}{405}=\frac{2}{81}\)
\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)
A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{100}\)
A : 3 = \(\frac{99}{100}\)
A = \(\frac{297}{100}\)
= 7/4.(4/1.5 + 4/5.9 + 4/9.13 + 4/13.17 + 4/17.21)
= 7/4.(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)
= 7/4.(1-1/21)
= 7/4.20/21 = 5/3
Tk mk nha
Đặt biểu thức bằng A
4/7A=1-1/5+1/5-1/9+...+1/17_1/21
4/7A=1-1/21
4/7A=20/21
A=35/21=5/3
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)
\(=1.\frac{15}{16}=\frac{15}{16}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)