B=\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)
Please help me!!!!!!!!
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Tính giá trị của biểu thức :\(\frac{1}{2\cdot4\cdot6}+\frac{1}{4\cdot6\cdot8}+\frac{1}{6\cdot8\cdot10}+...+\frac{1}{96\cdot98\cdot100}\)
Mong các bạn giúp đỡ !Thanks
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\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+.........+\frac{1}{99\cdot100}\)
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\(\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+\frac{3}{6\cdot8}+.....+\frac{3}{98\cdot100}\)
\(\frac{3}{2.4}+\frac{3}{4.6}+....+\frac{3}{98.100}\)
\(=\frac{3}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)
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\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)
\(=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{49}{100}=\frac{147}{200}\)
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Đặt \(A=\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{98.100}\)
\(A=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(A=\frac{3}{2}.\frac{49}{100}\)
\(A=\frac{147}{200}\)
Vậy \(A=\frac{147}{200}\)
Chúc bạn học tốt ~
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\(\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+.......+\frac{5}{100\cdot102}\)
help me , please everyone
Đặt \(A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
Ủng hộ mk nha !!! *_*
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\(\text{Đ}\text{ặt}:A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
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\(\frac{1}{1\cdot3}-\frac{1}{2\cdot4}+\frac{1}{3\cdot5}-\frac{1}{4\cdot6}+...+\frac{1}{97\cdot99}\frac{1}{98\cdot100}\)
tính hợp lí :
B=\(\frac{1\cdot4}{2\cdot3}+\frac{2\cdot5}{3\cdot4}+\frac{3\cdot6}{4\cdot5}+.....+\frac{98\cdot101}{99\cdot100}\)
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)=?
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{3}-\frac{1}{100}\)
= \(\frac{97}{300}\)
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\(E=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+....+\frac{1}{2016\cdot2018}\)
\(E=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2016.2018}\)
\(E=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2018-2016}{2016.2018}\)
\(2E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(E=\left(\frac{1}{2}-\frac{1}{2018}\right).\frac{1}{2}\)
\(E=\frac{504}{1009}.\frac{1}{2}\)
\(E=\frac{252}{1009}\)
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\(E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(E=\frac{1}{2}-\frac{1}{2018}\)
\(E=\frac{1005}{2018}\)
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\(E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2116}-\frac{1}{2018}\)
\(E=\frac{1}{2}-\frac{1}{2018}\)
\(E=\frac{1005}{2018}\)
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Tính \(\frac{1}{2\cdot4}\)+\(\frac{1}{4\cdot6}+...+\frac{1}{98\cdot100}\)
Đặt \(D=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)
=>\(2D=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)
=>\(2D=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)
=>\(2D=\frac{1}{2}-\frac{1}{100}\)
=>\(2D=\frac{49}{100}\)
=>\(D=\frac{49}{50}\)
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