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

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

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

\(=1-\frac{1}{11}=\frac{10}{11}\)

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

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

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

Bài 1: 

\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)

\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)

\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)

=1/131

Bài 2:

b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)

\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{3}{5}\)

`=`\(\dfrac{9}{12}+\dfrac{8}{12}+\dfrac{3}{5}\)

`=`\(\dfrac{17}{12}+\dfrac{3}{5}\)

`=`\(\dfrac{85}{60}+\dfrac{36}{60}\)

`=`\(\dfrac{121}{60}\)

`b)`

\(\dfrac{1}{2}\cdot\dfrac{9}{13}\div\dfrac{27}{26}\)

`=`\(\dfrac{1}{2}\cdot\dfrac{9}{13}\cdot\dfrac{26}{27}\)

`=`\(\dfrac{1}{2}\cdot\dfrac{2}{3}\)

`=`\(\dfrac{1}{3}\)

`c)`

\(\dfrac{2}{7}\cdot\dfrac{1}{9}+\dfrac{2}{7}\cdot\dfrac{2}{9}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{2}{7}\cdot\left(\dfrac{1}{9}+\dfrac{2}{9}\right)+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{2}{7}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{1}{3}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)

`=`\(\dfrac{1}{3}\cdot1=\dfrac{1}{3}\)

`d)`

\(11\div\dfrac{5}{2}+11\div\dfrac{7}{3}+11\div\dfrac{35}{6}\)

`=`\(11\cdot\dfrac{2}{5}+11\cdot\dfrac{3}{7}+11\cdot\dfrac{6}{35}\)

`=`\(11\cdot\left(\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{6}{35}\right)\)

`=`\(11\cdot1=11\)

a) 3/4 + 2/3 + 3/5 = 45/60 + 40/60 + 36/60 = 121/60

b) 1/2 x 9/13 : 27/26 = 9/26 x 26/27 = 1/3

c) 2/7 x 1/9 + 2/7 x 2/9 + 1/3 x 5/7 = 2/7 x (1/9 + 2/9) + 5/21 = 2/7 x 1/3 + 5/21 = 2/21 + 5/21 = 1/3

d) 11 : 5/2 + 11 : 7:3 + 11 : 35/6 = 11 x (2/5 + 3/7 + 6/35) = 11 x 1 = 11 

a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)

\(=\dfrac{9}{16}\)

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

\(B=0+0+0+0+0+\frac{9}{16}\)

\(B=0+\frac{9}{16}\)

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

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

\(B=0+0+0+0+0+\frac{9}{16}=\frac{9}{16}\)

Kết quả 1: Tính

Kết quả 2: Tính

\(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{11}+\dfrac{7}{13}-\dfrac{9}{16}-\dfrac{7}{13}+\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\\ =\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)+\left(-\dfrac{2}{11}+\dfrac{2}{11}\right)+\left(\dfrac{7}{13}-\dfrac{7}{13}\right)+\left(-\dfrac{9}{16}\right)\\ =0+0+0+0+0+\dfrac{-9}{16}\\ =\dfrac{-9}{16}=-0,5625\)

mik học lớp 5

Ta có : B=(1/5-1/5)-(3/7-3/7)+(5/9-5/9)-(2/11-2/11)+(7/13-7/13)-9/16

B=-9/16

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400