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\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)
A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)
A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
A=\(\dfrac{98}{99}-\dfrac{49}{100}\)
A=\(\dfrac{4949}{9900}\)
Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)
Vậy A<\(\dfrac{3}{4}\)
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)
\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)
\(x=\dfrac{198}{99}\)
Vậy \(x=\dfrac{198}{99}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{99^2}{98.100}\)
\(=\dfrac{2.2.3.3.4.4.....99.99}{1.3.2.4.3.5.....98.100}\)
\(=\dfrac{2.3.4.....99}{1.2.3.4.....98}.\dfrac{2.3.4.....99}{3.4.5.....100}\)
\(=\dfrac{99}{98}\cdot\dfrac{2}{100}\)
\(=\dfrac{99}{4900}\)