tính theo công thức bạn nhé

bằng 13/15 bạn ạ 

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

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

\(=0+0+0+0+0-\frac{11}{13}=-\frac{11}{13}\)

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

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

\(=0+0+0+0+0-\frac{11}{13}=-\frac{11}{13}\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

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

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

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

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

B=4*13/9*3-4/3*40/9

B=4/3*13/9-4/3*40/9

B=4/3*(13/9-40/9)

B=4/3*(-27)/9

B=4*(-3)/9

B=-4

A=6/7 + 1/7.(2/7+5/7)

A=6/7 + 1/7.7/7=6/7+1/7.1

A=6/7+1/7=7/7=1

a: \(\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-3^{10}\cdot5^{21}}{3^{12}\cdot5^{20}}=\frac{-5}{3^2}=-\frac59\)

b: \(\frac{-11^5\cdot13^7}{11^5\cdot13^8}=\frac{-11^5}{11^5}\cdot\frac{13^7}{13^8}=-\frac{1}{13}\)

c: \(2^{10}\cdot3^{10}-2^{10}\cdot3^9\)

\(=2^{10}\cdot3^9\cdot\left(3-1\right)\)

\(=2^{11}\cdot3^9\)

Ta có: \(\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)

\(=\frac{2^{11}\cdot3^9}{2^{10}\cdot3^9}=2\)

d: \(5^{11}\cdot7^{12}+5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(7+1\right)=5^{11}\cdot7^{11}\cdot8\)

\(5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(5\cdot7+9\right)\)

\(=5^{11}\cdot7^{11}\left(35+9\right)=5^{11}\cdot7^{11}\cdot44\)

Ta có: \(\frac{5^{11}\cdot7^{12}+5^{11}\cdot7^{11}}{5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}}\)

\(=\frac{5^{11}\cdot7^{11}\cdot8}{5^{11}\cdot7^{11}\cdot44}=\frac{8}{44}=\frac{2}{11}\)

Hicc mình hông còn tgian mấy bạn giúp minh 2 câu thui cũng đc ạk

a)  5/9 + 4/9 . 3/7 + 4/9 . 4/7

  = 5/9 + 4/9 . (3/7 + 4/7)

  = 5/9 + 4/9 . 1

  = 5/9 + 4/9

  = 1