Question

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

Answer 1

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

\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

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

\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)

Answer 2

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

=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)

=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)

=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

=\(2\left(1-\frac{1}{1005}\right)\)

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

=\(\frac{2008}{1005}\)

P/s: Không biết đúng không nữa, làm đại ^.^

Answer 3

Ta thấy : \(\frac{4}{2.4}=\frac{1}{2}\left(\frac{4}{2}-\frac{4}{4}\right);\frac{4}{4.6}=\frac{1}{2}\left(\frac{4}{4}-\frac{4}{6}\right);...;\frac{4}{2008.2010}=\frac{1}{2}\left(\frac{4}{2008}-\frac{4}{2010}\right)\)

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

=> C = \(\frac{1}{2}\left(\frac{4}{2}-\frac{4}{2010}\right)=\frac{1}{2}\left(2-\frac{2}{1005}\right)=\frac{1}{2}\left(\frac{2010}{1005}-\frac{2}{1005}\right)\)

=> C = \(\frac{1}{2}\left(\frac{2010-2}{1005}\right)=\frac{1}{2}.\frac{2008}{1005}=\frac{1004}{1005}\)

 

Answer 4

... một bài toán có 3 kết quả........ ôi zời...........

Answer 5

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

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

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

\(C=2.\frac{502}{1005}\)

\(C=\frac{1004}{1005}\)

Answer 6

 Sai hết rùi

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

C=\(\frac{4}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{1004}{1005}\)

,

Answer 7

khó hỉu thiệt;-;"