\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)

\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{25}{152}-\frac{4}{27}\)

\(C=\frac{67}{4104}\)

ra 67/1014

anh gửi chơi hay thật vậy

a)b) Bạn nhân cả tử và mẫu với 2. Mình làm luôn, ko ghi lại đề bài

a)\(\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)

=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)

=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)

=\(\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)

b)\(\frac{2}{10.18}+\frac{2}{18.26}+\frac{2}{26.34}+...+\frac{2}{802.810}\)

=\(\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\right)\)

=\(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{810}\right)\)

=\(\frac{1}{4}.\frac{8}{81}=\frac{2}{81}\)

c) Mình biết làm, ddoiwtj tí nữa mình làm cho. Giờ đang mỏi tay

Thẳng Nobita kun có chép bài thì đừng t..i..c..k cho nó

Đề bài của bạn thật là khó hiểu 

A=\(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

 =\(\frac{1}{2}-\frac{1}{509}\)

 =\(\frac{507}{1018}\)

sory em học lớp 5 không biết làm nếu biết em đã làm rồi hihihih.....

Có ai trả lời đi

\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)

\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)

\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)