Ôn tập chương III

ND

\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)

\(X=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}\)

\(K=\dfrac{1}{3}.\dfrac{1}{15}.\dfrac{1}{35}...\dfrac{1}{9999}\)

H24
9 tháng 4 2017 lúc 18:37

\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)

\(\Rightarrow G=\dfrac{64}{505}\)

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ND
9 tháng 4 2017 lúc 9:39

giải hộ với

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HL
9 tháng 4 2017 lúc 9:45

\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\\ G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\\ G=\dfrac{2}{3}.\dfrac{96}{505}\\ G=\dfrac{64}{505}\)

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HL
9 tháng 4 2017 lúc 9:50

\(X=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}\\ X=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}...\dfrac{99.99}{98.100}\\ X=\dfrac{2.3.4...99}{1.2.3....98}.\dfrac{2.3.4...99}{3.4.5...100}\\ X=\dfrac{99}{1}.\dfrac{2}{100}\\ X=\dfrac{99}{50}\)

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DK
13 tháng 4 2017 lúc 11:13

G= \(\dfrac{2}{3}.\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{98.101}\right)\)

G=\(\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

G= \(\dfrac{2}{3}-\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

G= \(\dfrac{2}{3}-\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\)

G=\(\dfrac{2}{3}.\dfrac{96}{505}\)

G= 192/1515

G=64/505

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ND
14 tháng 4 2017 lúc 20:26

giải câu k đi

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NQ
11 tháng 5 2017 lúc 19:36

\(G=\dfrac{2}{5\cdot8}+\dfrac{2}{8\cdot11}+...+\dfrac{2}{95\cdot98}+\dfrac{2}{98\cdot101}\)

\(=\dfrac{2}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{95\cdot98}+\dfrac{3}{98\cdot101}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{96}{505}\)

\(=\dfrac{64}{505}\)

\(X=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}...\dfrac{99^2}{98\cdot100}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}...\dfrac{99\cdot99}{98\cdot100}\)

\(=\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4...99\cdot99}{1\cdot3\cdot2\cdot4\cdot3\cdot5...98\cdot100}\)

\(=\dfrac{2\cdot99}{100}=\dfrac{198}{100}=\dfrac{99}{50}\)

\(K=\dfrac{1}{3}\cdot\dfrac{1}{15}\cdot\dfrac{1}{35}...\dfrac{1}{9999}\)

\(=\dfrac{1}{1\cdot3}\cdot\dfrac{1}{3\cdot5}\cdot\dfrac{1}{5\cdot7}...\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{1\cdot3\cdot3\cdot5\cdot5\cdot7...99\cdot101}\)

\(=\dfrac{1}{3^2\cdot5^2\cdot7^2...99^2\cdot101}\)

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