\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\...">
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23 tháng 3 2018

mình làm được nhưng đánh lâu lắm

6 tháng 3 2018

em lop 4 ma???????????????????????

6 tháng 7 2017

= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101

= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101

= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)

= 100.2/101

=200/101

7 tháng 3 2018

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)

\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)

\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)

\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)

2 tháng 5 2017

Ta có

=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{8.10}\right)\)

=\(\frac{4}{3}.\frac{9}{8}....\frac{81}{80}\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{9.9}{8.10}\)

=\(\frac{2.3....9}{1.2....8}.\frac{2.3....9}{3.4....10}\)

=\(9.\frac{2}{10}\)

=\(\frac{9}{5}\)

2 tháng 4 2015

Ta có : A = (4/1.3) . (9/2.4).......(10000/99.101)

               = (2.2/1.3). (3.3/2.4).......(100.100/99.101)

              =(2.3.4......99.100/1.2.3.....98.99 ) . ( 2.3.4.......100/3.4.5.....101)

               =(100/1) . ( 2/101 )

              =200/101

26 tháng 8 2015

a, Đúng rồi đó

b, \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+....+\left(1+\frac{1}{39.41}\right)\)

\(\frac{4}{1.3}.\frac{9}{2.4}.....\frac{1600}{39.41}\)

\(\frac{2.2.3.3....40.40}{1.3.2.4....39.41}\)

\(\frac{\left(2.3....40\right)\left(2.3....40\right)}{\left(1.2....39\right)\left(3.4....41\right)}\)

\(\frac{40.2}{41}\)

\(\frac{80}{41}\)

26 tháng 8 2015

\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{8^2}\right)\)

\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{8^2-1}{8^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{7.9}{8.8}\)

\(=\frac{\left(1.2.....7\right).\left(3.4.....9\right)}{\left(2.3.....8\right).\left(2.3.....8\right)}\)

\(=\frac{1.9}{8.2}=\frac{9}{16}\)

14 tháng 4 2019

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\Rightarrow T=\frac{1004}{1005}\)

14 tháng 4 2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)