Question

Tính các tổng sau  

A = 3/4.7 + 4/7.11 + 4/11.15 + ... + 4/100.104

B = 1/25.27 + 1/27.29 + 1/29.31 +...+ 1/73.75

C = 6/15.18 + 6/18.21 + 6/21.24 +...+6/87.90

Các ban ơi dấu chấm là dấu nhân nhé nhớ giải giúp mình nhé mình thank you các bạn nhiều lắm

Answer 1

\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)

\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)

\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{4}-\frac{1}{104}\)

\(A=\frac{25}{104}\)

Answer 2

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)

\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)

\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)

\(B\cdot2=\frac{2}{75}\)

\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)

\(B=\frac{1}{75}\)

Answer 3

\(C=\frac{6}{15\cdot18}+\frac{6}{18\cdot21}+\frac{6}{21\cdot24}+...+\frac{6}{87\cdot90}\)

\(\frac{C}{2}=\frac{3}{15\cdot18}+\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+...+\frac{3}{87\cdot90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{18}\)

\(C=\frac{1}{18}\cdot2\)

\(C=\frac{1}{9}\)

Answer 4

dấu chấm là dấu nhân à, vậy dấu ... là 3 dấu nhân đúng ko :>>

Answer 5

A=\(\frac{3}{4.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)

Đặt E=\(\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)

       E=\(\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

       E=\(\frac{1}{7}-\frac{1}{104}\)

\(\rightarrow\)A=\(\frac{3}{4.7}\)+E

         A=\(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{104}\)

         A=\(\frac{1}{4}-\frac{1}{104}\)

         A=\(\frac{25}{104}\)

Vậy A=\(\frac{25}{104}\)

B=\(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

B=\(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)+\frac{1}{2}\left(\frac{1}{27}-\frac{1}{29}\right)+...+\frac{1}{2}\left(\frac{1}{73}-\frac{1}{75}\right)\)

B=\(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

B=\(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

B=\(\frac{1}{2}.\frac{2}{75}\)

B=\(\frac{1}{75}\)

Vậy B=\(\frac{1}{75}\)

Câu C tương tự