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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)

\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)

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

\(A=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

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

\(=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

Ta thấy :

 \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

\(.........\)

\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)

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

\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)

đáp án = \(\frac{297}{100}\)

đúng không?

kết bạn với mh nha

`2,86 . 4 + 3,14 . 4 – 6,01 . 5 + 32. 0,75`

`=4.(2,86+3,14)-30,05+24`

`=4.6-30,05+24`

`=24-30,05+24`

`=17,95`

\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

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

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

\(A=\dfrac{3^2}{1\times4}+\dfrac{3^2}{4\times7}+\dfrac{3^2}{7\times10}+\dfrac{3^2}{10\times13}+\dfrac{3^2}{13\times16}...+\dfrac{3^2}{97\times100}\)

\(=3\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16} +...+\dfrac{3}{97\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

\(=2\dfrac{97}{100}\)

Vậy \(A=2\dfrac{97}{100}\)

Tổng hai số bằng 1902.biết số thứ nhất lớn hơn số thứ hai.196 đơn vị .tìm hai số đó cô cho đề bài khó quá nghỉ hè quên hết kiến thức lớp 4

Mình cần rất gấp cực kì gấp

mình học cấp 2 bạn ạ

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)

\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{255}{128}+\frac{3}{512}\)

\(=\frac{1020}{512}+\frac{3}{512}\)

\(=\frac{1023}{512}\)

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
                                                        \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
                                                         \(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)