Question

[Nhật Mai]

HSBT toán 2k9 câu đánh dấu thui nhé

Answer 1

[Nguyễn Minh Quang]

e.\(E=\frac{1}{19}+\frac{1}{19.29}+\frac{1}{29.39}+..+\frac{1}{2009.2019}\)

\(=\frac{1}{19}+\frac{1}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+..+\frac{1}{2009}-\frac{1}{2019}\right)=\frac{1}{19}+\frac{1}{10}\left(\frac{1}{19}-\frac{1}{2019}\right)\)

\(=\frac{1}{19}+\frac{200}{19\times2019}=\frac{2219}{19\times2019}\)

g.\(G=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)

\(H=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{10.11.12}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{65}{264}\)

\(\)câu I thì mình không nhìn thấy chỗ ... sẽ là gì nên không thể tính được

Answer 2

[Nguyễn Minh Quang]

à cảm ơn bạn để mình làm luôn câu i

\(I=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}\right)\)

\(=100\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}\right)=100\left(1-\frac{1}{100}\right)=99\)