Vậy \(A=\frac{8056}{2015}\)

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\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)

\(\frac{1}{4}A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)

\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)

\(\frac{1}{4}A=1-\frac{1}{2015}\)

\(\frac{1}{4}A=\frac{2014}{2015}\)

\(A=\frac{2014}{2015}:\frac{1}{4}=\frac{2014}{2015}\cdot4=\frac{8056}{2015}\)

Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.

Đó là số 88.

Xem thêm tại: http://loigiaihay.com/bai-99-trang-39-sgk-toan-6-tap-1-c41a3896.html#ixzz4xczZ4dOb

\(B=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{29.30}\right)\)

\(B=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{29}-\frac{1}{30}\right)\)

\(B=4.\left(1-\frac{1}{30}\right)\)

\(B=4.\frac{29}{30}\)

\(B=\frac{58}{15}\)

\(B=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{29}-\frac{1}{30}\right)\) 

\(=4\left(1-\frac{1}{30}\right)\) 

\(=4.\frac{29}{30}=\frac{58}{15}\) 

Vậy B= \(\frac{58}{15}\)

So sánh à bạn?

A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)

B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)

vậy A=B

\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)

\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)

\(\Rightarrow A=B\)

2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)

  \(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\frac{2575}{5151}\)

  \(=2,499514657\)

= 2,499514657 bạn nhé

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)

b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)\)

\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2.\frac{2018}{2019}\)

\(=\frac{4036}{2019}\)

Phần c tương tự nha

a) \(\frac{1}{1.2}\) +  \(\frac{1}{2.3}\) + .......+  \(\frac{1}{2017.2018}\)

= 1 -  \(\frac{1}{2}\) + \(\frac{1}{2}\) -  \(\frac{1}{3}\) + .......+  \(\frac{1}{2017}\) -   \(\frac{1}{2018}\)

= 1 -  \(\frac{1}{2018}\) =  \(\frac{2017}{2018}\)

câu a) mik sửa đề một tí ko biết có đúng ko

câu b , c tương tự nhưng cần lấy tử ra chung 

a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)nhóm 2 ra ngoài rồi làm như câu a

c)nhóm 4 ra rồi làm như câu a

#)Giải :

Đặt \(A=4-\frac{2}{1.2}-\frac{2}{2.3}-\frac{2}{3.4}-...-\frac{2}{99.100}\)

\(A=4-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)

\(A=4-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=4-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=4-2\left(1-\frac{1}{100}\right)\)

\(A=4-2\times\frac{99}{100}\)

\(A=4-\frac{99}{50}\)

\(A=\frac{101}{50}\)