Ta thấy: \(1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(\Rightarrow A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)

\(=\left(\frac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\right)\left(\frac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}\right)\left(\frac{\left(4-1\right)\left(4+2\right)}{4\left(4+1\right)}\right)...\left(\frac{\left(2006-1\right)\left(2006+2\right)}{2006\left(2006+1\right)}\right)\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2005.2008}{2006.2007}=\frac{\left(1.2.3...2005\right)\left(4.5.6...2008\right)}{\left(2.3.4...2006\right)\left(3.4.5...2007\right)}\)

\(=\frac{1.2008}{2006.3}=\frac{1004}{1003.3}=\frac{1004}{3009}\)

Vậy \(A=\frac{1004}{3009}\)

Lời giải:

Xét công thức tổng quát:

$1+2+3+...+n=\frac{n(n+1)}{2}$

$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$

Thay $n=2,3,...,2006$ ta thu được:

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)

\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)

Lời giải:

Xét công thức tổng quát:

$1+2+3+...+n=\frac{n(n+1)}{2}$

$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$

Thay $n=2,3,...,2006$ ta thu được:

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)

\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)

\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+2006})\)

\(A=(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{1}{2013021})\)

\(A=\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}....\frac{2013020}{2013021}\)

Sorry bạn máy tính mình có chút vấn đề để mk làm tiếp :

\(A=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}....\cdot\frac{4026040}{4026042}\)

\(A=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{2005\cdot2008}{2006\cdot2007}\)

\(A=\frac{1\cdot2\cdot3\cdot...\cdot2005}{2\cdot3\cdot4\cdot...\cdot2006}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2008}{3\cdot4\cdot5\cdot...\cdot2007}\)

\(A=\frac{1}{2006}\cdot\frac{2008}{3}=\frac{1004}{3009}\)

P/S : Hoq chắc :>

dsf

\(\left(-\frac{1}{2}\right)+\left(-\frac{1}{9}\right)-\left(-\frac{3}{5}\right)+\frac{1}{2006}-\left(-\frac{2}{7}\right)\)

\(=\left(-\frac{1}{2}\right)-\frac{1}{9}+\frac{3}{5}+\frac{1}{2006}+\frac{2}{7}\)

\(=\left[\left(-\frac{1003}{2006}+\frac{1}{2006}\right)\right]+\left[\left(-\frac{5}{45}+\frac{27}{45}\right)\right]+\frac{2}{7}\)

\(=-\frac{1002}{2006}+\frac{22}{45}+\frac{2}{7}\)

\(=-\frac{501}{1003}+\frac{154}{315}+\frac{90}{315}\)

\(=-\frac{501}{1003}+\frac{244}{315}\)

\(=-\frac{157815}{315945}+\frac{244732}{315945}=\frac{86917}{315945}\approx0,28\)

P/s : Vì bài này số quá xấu nên mình đổi ra số thập phân cho gọn nhé ....

\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+...+2006}\right)\)

\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{2013021}\right)\)

\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}....\frac{2013020}{2013021}\)

\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}......\frac{4026040}{4026042}\)

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}......\frac{2005.2008}{2006.2007}\)

\(A=\frac{1.2.3.....2005}{2.3.4....2006}.\frac{4.5.6....2008}{3.4.5...2007}\)

\(A=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)

Ta có: 

\(1-\frac{1}{1+2}=1-\frac{1}{2.3:2}=1-\frac{2}{6}=\frac{4}{6}=\frac{1.4}{2.3}\)

\(1-\frac{1}{1+2+3}=1-\frac{1}{3.4:2}=1-\frac{2}{12}=\frac{10}{12}=\frac{2.5}{3.4}\)

\(1-\frac{1}{1+2+3+4}=1-\frac{1}{4.5:2}=1-\frac{2}{20}=\frac{18}{20}=\frac{3.6}{4.5}\)

...

\(1-\frac{1}{1+2+3+...+2006}=1-\frac{1}{2006.2007:2}=1-\frac{2}{2006.2007}=\frac{2005.2008}{2006.2007}\)

=> \(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{2005.2008}{2006.2007}\)

\(=\frac{\left(1.2.3....2005\right).\left(4.5.6...2008\right)}{\left(2.3.4...2006\right)\left(3.4.5...2007\right)}=\frac{1}{2006}.\frac{2008}{3}=\frac{2008}{6018}\)

Một mảnh vườn hình chữ nhật có nữa chu vi 120m , chiều dài hơn chiều rộng 20m .

a) Tính diện tích mảnh vườn ình chữ nhật đó ?

b) Người ta dùng 4/7 diện tích đó đẻ trồng hoa . Hỏi diện tích trồng hoa là bao nhiêu mét vuông ?