A=\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)

3A=3(\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\))

3A=\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

3A=\(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}+\frac{1}{76}\)

3A=\(\frac{1}{4}-\frac{1}{76}\)

3A=\(\frac{9}{38}\)

A=\(\frac{9}{38}\):3

A=\(\frac{3}{38}\)

đặt A=1/4.7+1/7.10+...+1/73.76

3A=1/4-1/7+1/7-1/10+...+1/ 73 -1/ 76
 

\(A=\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)

\(3A=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(3A=\frac{1}{4}-\frac{1}{76}\)

\(3A=\frac{9}{38}\)

\(A=\frac{3}{38}\)

Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)

           1/4-1/7 = 3/28 = 3.(1/4.7)

A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)

A = 3.(1-1/100)

A = 3.(99/100)

A = 297/100

\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\frac{99}{100}\)

\(A=\frac{33}{100}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)

\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}.\frac{18}{19}\)

\(A=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)

\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}.\frac{5}{24}\)

\(B=\frac{5}{48}\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)=\frac{1}{3}.\frac{18}{19}=\frac{6}{19}\) VẬY: \(A=\frac{6}{19}\)

\(B=\frac{1}{4.2.4}+\frac{1}{4.4.6}+...+\)\(\frac{1}{4.10.12}\)

\(B=\frac{1}{4}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{10.12}\right)\)\(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{10}-\frac{1}{12}\right)\)

                                                                               \(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{120}\right)=\frac{1}{8}.\frac{59}{120}=\frac{59}{960}\)

đây là toán lớp 5 cơ mà

a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)

A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)

A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))

A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)

A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))

A=\(\frac{1}{3}\)x\(\frac{18}{19}\)

A=\(\frac{6}{19}\)

câu b tương tự tách mẫu ra thôi

mik ko hiểu cách làm của bạn cho lắm ,có thể làm rõ hơn đc ko ???

http://olm.vn/hoi-dap/question/772291.html

sau 3 phút có kết quả tùy bạn

A=\(\frac{37}{78}\)

B=\(\frac{9}{38}\)

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 

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