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\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{5047}{30300}\right)\Rightarrow S=\frac{5047}{12120}\)
Tính:
S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2S=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4949}{9900}\)
\(S=\frac{4949}{19800}\)
Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
...
\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)
=> 2S = \(\frac{4949}{9900}\)
=> S = \(\frac{4949}{19800}\)
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}=\frac{3.2.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
\(S:3.2=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{98.99.100}\)
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
Tương tự nhé ta có
\(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
\(S=\frac{4949}{6600}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{101-99}{99.100.101}\)
\(C=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}+\frac{2}{100.101}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(C=\frac{1}{2}\cdot\frac{5049}{10100}=\frac{5049}{20200}\)
Bài này hơi dài nên bạn tham khảo tại đây nha :
Câu hỏi của Kim Sura xXx pÉ heO - Toán lớp 6 - Học toán với OnlineMath
làm tiếp theo
\(S=\frac{5}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}+\frac{2}{99.100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
còn lại tự làm
\(S=\frac{5}{2\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+......+\frac{5}{99\cdot100\cdot101}\)
\(S\frac{2}{5}=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+.....+\frac{2}{99\cdot100\cdot101}\)
\(\frac{2}{2\cdot3\cdot4}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\)
\(\frac{2}{3\cdot4\cdot5}=\frac{1}{3\cdot4}-\frac{1}{4\cdot5}\)
.............
\(\frac{2}{99\cdot100\cdot101}=\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.........+\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{6}-\frac{1}{10100}\)
\(\Rightarrow S\frac{2}{5}=\frac{5047}{30300}\)
\(\Rightarrow S=\frac{5047}{30300}:\frac{2}{5}\)
\(\Rightarrow S=\frac{5047}{30300}\cdot\frac{5}{2}\)
\(\Rightarrow S=\frac{5047}{12120}\)