Question

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

tính c

Answer 1

TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH

                               STUDY   WELL

      K NHA

MK XIN CẢM ƠN CÁC BẠN NHÌU

Answer 2

C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405 

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

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

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

\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)

 \(C=\frac{25}{152}-\frac{4}{27}\)

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

Study well 

Answer 3

\(C=\frac{2}{4\cdot7}-\frac{3}{5\cdot9}+\frac{2}{7\cdot10}-\frac{3}{9\cdot13}+...+\frac{2}{301\cdot304}-\frac{3}{401\cdot405}\)

\(C=\left(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{301\cdot304}\right)-\left(\frac{3}{5\cdot9}+\frac{3}{9\cdot13}+...+\frac{3}{401\cdot405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{301\cdot304}\right)-\frac{3}{4}\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{401\cdot405}\right)\)

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

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

\(C=\frac{2}{3}\cdot\frac{77}{304}-\frac{3}{4}\cdot\frac{82}{405}\)

\(C=\frac{77}{456}-\frac{41}{270}\)

\(C=\frac{349}{20520}\)

Không chắc =))

Answer 4

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

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

Đặt D = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\)

      \(=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)

        \(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\right)\)

          \(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)\)

            \(=\frac{2}{3}.\frac{75}{304}=\frac{25}{152}\)

Đặt E = \(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\)

\(=\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)

...Đoạn tiếp làm tương tự như D

\(\Rightarrow E=\frac{4}{27}\)

Thay D và E vào C  có :

C = \(\frac{25}{152}-\frac{4}{27}=\frac{67}{4104}\)