A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2011 - 1/2012

A = 1 - 1/2012

A = 2011/2012

B = 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +...+ 1/2010 - 1/2012

B = 1/2 - 1/2012

B = 1005/2012

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(A=1-\dfrac{1}{2012}\)

\(A=\dfrac{2011}{2012}\)

b) \(B=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2010\cdot2012}\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2010\cdot2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2010}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2012}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{1005}{2012}\)

\(B=\dfrac{1005}{4024}\)

a, \(A=\frac{6}{10.11}+\frac{6}{11.12}+\frac{6}{12.13}+...+\frac{6}{69.70}\)

\(A=\frac{6}{10}-\frac{6}{11}+\frac{6}{11}-\frac{6}{12}+\frac{6}{12}-\frac{6}{13}+...+\frac{6}{69}-\frac{6}{70}\)

\(A=\frac{6}{10}-\frac{6}{70}\)

\(A=\frac{18}{35}\)

b, \(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)

\(B=\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(B=2.\frac{1009}{2020}\)

\(B=\frac{1009}{1010}\)

Chúc bạn học tốt

a)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)

\(=\frac{1\cdot2\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{3\cdot4\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\)

\(=\frac{1\cdot2}{3\cdot4}=\frac{1}{6}\)

b) \(\frac{12\cdot13+24\cdot26+36\cdot39}{24\cdot26+48\cdot52+72\cdot78}\)

\(=\frac{12\cdot13\cdot\left(1+2\cdot2+3\cdot3\right)}{24\cdot26\cdot\left(1+2\cdot2+3\cdot3\right)}\)

\(=\frac{12\cdot13}{24\cdot26}=\frac{1}{4}\)

\(\frac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\frac{2\cdot3\cdot\left(1+2\cdot2+7\cdot7\right)}{3\cdot5\cdot\left(1+2\cdot2+7\cdot7\right)}=\frac{2\cdot3}{3\cdot5}=\frac{2}{5}\)

=3*(1/1.2+1/2.3+...+1/2018.2019)

=3(1-1/2+1/2-1/3+...+1/2018-1/2019)

=3(1-1/2019)

=3*2018/2019

=2018/673

\(A=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{2018.2019}\)

  \(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)

   \(=3.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

    \(=3.\left(1-\frac{1}{2019}\right)\)

     \(=3.\frac{2018}{2019}=\frac{2018}{673}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)

\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)

\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)

\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)

a, 1/1.2+1/2.3+1/3.4+...+1/999.1000

=  1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000

=   1/1-1/1000

=   999/1000

b, 1/2.4+1/4.6+1/6.8+1/8.10

=  1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10

=  1/2-1/10

=   4/10  =2/5

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9.

Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)

\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)

\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)

\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

\(S=\dfrac{1009}{2019}\)

Còn lại bạn làm tương tự hết nhé .

Gợi ý 

phân số tối giản có ước chung lớn nhất là 1

tim d (UCLN 4n+1;12n+7)

ta có d= +-1;+-2;+-4

sau đó CM 4n+1 ko chia hết cho 2 và 4 và ta tìm đc d=1

b) Đầu tiên lấy cả cụm A trừ đi cụm B phá ngoặc rồi nhóm 1.2-1.3;2.3-2.4...

rồi công -1+(-2)+(-3)+...+(-2016)

ta tìm đc kết quả là -2033136

à đúng rồi, mk cần các bạn giải chi tiết bài giải giùm mk! ;)