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13 tháng 3 2020

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

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2010}\)

\(M=1-\frac{1}{2010}=\frac{2009}{2010}\)

Vậy M=\(\frac{2009}{2010}\)

+) Đặt A=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{50}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\frac{49}{50}\)

\(A=\frac{1\cdot2\cdot\cdot\cdot\cdot49}{2\cdot3\cdot\cdot\cdot\cdot50}=\frac{1}{50}\)

8 tháng 3 2022

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

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

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

21 tháng 1 2022

làm chi tiết đc ko ạ

28 tháng 3 2017

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

5 tháng 5 2022

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:

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

\(=>x=3\)

14 tháng 6 2020

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

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

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

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

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

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 \(B=\frac{4949}{19800}.\)

14 tháng 6 2020

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

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

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

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

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

23 tháng 5 2017

sorry mình nhầm

ta có:

M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)

=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)

=\(\frac{1}{5}\)

vậy M=\(\frac{1}{5}\)

23 tháng 5 2017

\(M=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

6 tháng 5 2017

\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot\frac{5^2}{5\cdot6}=\frac{1^2}{1\cdot6}=\frac{1}{6}\)

lan sau nho ghi de cho dung nha bn

6 tháng 5 2017

\(\frac{1.1.2.2.3.3.4.4.5.5}{1.2.2.3.3.4.4.5.5.6}\)=\(\frac{\left(1.2.3.4.5\right).\left(1.2.3.4.5\right)}{\left(1.2.3.4.5\right)\left(2.3.4.5.6\right)}=\frac{1}{6}\)

22 tháng 7 2016

\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)

\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)

\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)

                      \(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)

                      \(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)

\(\Rightarrow B=\frac{B}{\frac{2}{B}}=\frac{1}{2}\)