Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Ta có:

\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)

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

\(S=\dfrac{2017}{2018}\)

=1/1.2+1/2.3+1/3.4+...1/2017.2018

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2018/2018-1/2018

=2017/2018

S = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ .......+ \(\dfrac{1}{2017.2018}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+.......+ \(\dfrac{1}{2017}\) - \(\dfrac{1}{2018}\)

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

S = \(\dfrac{2017}{2018}\)

sưả đề \(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}\)

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

\(P=1+\dfrac{-1}{100}\)

\(P=\dfrac{99}{100}\)

=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1-1/100

=99/100

Xét mẫu số:   1/(2x3) + 1/(3x4) + …… + 1/(99x100)

       = 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100

       = (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)

       = (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2

       = (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2  + 1/3 + ....... +1/50 )

       = 1/51 + 1/52 + 1/53 + ............. + 1/100            (Đơn giản số trừ)

         =>(1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50

= 49/50

ỦNG HỘ NHA

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)

 \(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

1/1.2+1/2.3+1/3.4+...+1/2017.2018

=1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2017/2018

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

Ta có công thức :

\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)

\(A=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)

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

\(=1-0-0-0-...-0-\frac{1}{n}\)

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

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)

\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)

Vậy \(A=\frac{2019}{505}.\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)

\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Vậy \(B=\frac{4949}{19800}.\)

\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)

\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)

\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)

Đến đây tự tính

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

Số hơi bị dữ nên tính nốt nhé

a) \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+........+\frac{4}{2019.2020}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2019.2020}\right)\)

\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=4.\left(1-\frac{1}{2020}\right)=4.\frac{2019}{2020}=\frac{2019}{505}\)