\(\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)

\(=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=\frac{-9}{16}\)

a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

A=0+0+0+...+0+\(\dfrac{13}{15}\)

A=\(\dfrac{13}{15}\)

b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)

\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}-\dfrac{9}{11}-\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\left(+\dfrac{7}{9}\rightarrow-\dfrac{7}{9}\right)\)

\(\Rightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=\dfrac{-11.15+13.13}{13.15}\)

\(\Rightarrow A=\dfrac{-165+169}{195}=\dfrac{4}{195}\)

Chắc bạn gõ nhầm số hạng thứ 3 phải là +5/7.

Tổng này đối xứng qua 13/15. Các đối xứng trái dấu nên Tổng = 13/15

Tính theo công thức nha bạn

tính theo công thức bạn nhé

1^3-3^5-(-3^5)+1^64-2^9-1^36+1^15

=1+(-3^5+3^5)+1-2^9-1+1

=2-2^9

=-510

bằng 13/15 bạn ạ 

a=-11/13

các bạn cho mình xin lời giải cụ thể với ạ

\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)

\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)

\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)