a=1.2 + 2.3 +3.4+ ...+ 2010.2011\

3a=1.2.3+2.3.3+3.4.3+......+2010.2011.3

3a=1.2.3+2.3.(4-1)+3.4.(5-1)+............+2010.2011.(2012-2009)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2010.2011.2012-2009.2010.2011

3a=2010.2011.2012

a=2010.2011.2012:3

a=?

A=2010.2011.2012:3

\(\frac{1.2+1.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)

=\(\frac{1.2}{2.3}\)+\(\frac{1.4}{4.6}\)+\(\frac{3.6}{6.9}\)+\(\frac{4.8}{8.12}\)

= \(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)

= \(\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{2}{6}\)

=\(\frac{7}{6}\)

Mình nghĩ đề bài phải là:

          \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)   *2.3 + 4.6 + 6.9 + 8.12 = 3.(1.2 + 2.4 + 3.6 + 4.8)*

\(=\)\(\frac{1\left(1.2+2.4+3.6+4.8\right)}{3\left(1.2+2.4+3.6+4.8\right)}\)

\(=\)\(\frac{1}{3}\)

A= ​​\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2005.2006}\)= \(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2005}\)-\(\frac{1}{2006}\)=

= 1-\(\frac{1}{2006}\)= \(\frac{2005}{2006}\)

a)Ta có:\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(\Rightarrow A=\frac{2005}{2006}\)

b)Ta có:\(\frac{2005}{2006}-1=-\frac{1}{2006}\)

        Vì \(\frac{2005}{2006}\) trừ 1 được một số âm thì chứng tỏ \(\frac{2005}{2006}\)<1

Vậy A<1

\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)

\(M=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)

\(M=\frac{1.2.3.....10}{1.2.3....10}.\frac{1.2.3.....10}{2.3.4.....11}\)

\(M=1.\frac{1}{11}\)

\(M=\frac{1}{11}\)

Cảm ơn bạn nha

A=1.2+2.3+3.4+4.5+5.6+...+2016.2017 

=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017

=> 3A = 2016.2017.2018

=> A = 2016.2017.2018 : 3 

Mày.ngu qua

Sửa đề: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

Đặt \(B=1\cdot98+2\cdot97+\cdots+98\cdot1\)

\(=2\left(1\cdot98+2\cdot97+\cdots+49\cdot50\right)\)

\(=2\cdot\left\lbrack1\left(99-1\right)+2\left(99-2\right)+\cdots+49\left(99-49\right)\right\rbrack\)

\(=2\cdot\left\lbrack99\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack99\cdot\frac{49\cdot50}{2}-\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack=2\cdot\left\lbrack99\cdot49\cdot25-49\cdot25\cdot33\right\rbrack\)

\(=2\cdot25\cdot49\cdot33\left(3-1\right)=50\cdot49\cdot33\cdot2=100\cdot49\cdot33\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+98\cdot99\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+98\left(98+1\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+\left(1+2+\cdots+98\right)\)

\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}+\frac{98\cdot99}{2}=49\cdot33\cdot197+49\cdot99=49\cdot33\cdot\left(197+3\right)=49\cdot33\cdot200\)

Ta có: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

\(=\frac{100\cdot49\cdot33}{49\cdot33\cdot200}=\frac{100}{200}=\frac12\)

Bạn tham khảo tại link này nhé :

Câu hỏi của Đỗ Minh Hùng - Toán lớp 6 - Học toán với OnlineMath

Đặt E = 1.2+2.3+3.4+4.5+ .... + 1001.1002

3E = 1.2+2.3+3.4+4.5+ .... + 1001.1002.3

3E = 1.2 (3 - 0) + 2.3.(4 - 1) + 3.4 . (5 - 2).... . 1001.1002. (1001 - 1000)

3E = (1.2.3 + 2.3.4 + 3.4.5 +....+ 1001.1002.1001) - (0.1.2 + 1.2.3 + 2.3.4 +....+ 1000.1001.1002)

3E = 1001.1002.1001 - 0.1.2

3E = 1004005002 - 0

3E = 1004005002

3E = 1004005002 : 3 

3E = 334668334

Không đúng thì vô đây kham khảo nhé.

Câu hỏi của lê trần hồng thắm - Toán lớp 6 - Học toán với OnlineMath

Ta có:

\(\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\) =\(\frac{2.3^2}{5}\) =\(\frac{18}{5}\)

Vậy

Ta cã: \(\frac{2^3.3^4}{2^2.3^2.5}\)

\(=\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\)

\(=\frac{2.3^2}{5}\)

\(=\frac{18}{5}\)