\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).................\left(1-\frac{1}{999}\right).\left(1-\frac{1}{1000}\right)\)

\(P=\frac{-1}{2}.\frac{-2}{3}.......................\frac{-998}{999}.\frac{-999}{1000}\)

\(P=\frac{\left(-1\right).\left(-2\right)...............\left(-998\right).\left(-999\right)}{2.3........................999.1000}\)

\(P=\frac{-1}{1000}\)

thank you bạn nha

\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)

Vậy B = - 2016

Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{999}\right)\left(1-\frac{1}{1000}\right)\)

\(P=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{998}{999}\cdot\frac{999}{1000}\)

\(P=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot999}{2\cdot3\cdot4\cdot5\cdot...\cdot1000}\)

\(P=\frac{1}{1000}\)

\(P=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{998}{999}\times\frac{999}{1000}\)

P=1/1000

_Kudo_

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot.....\cdot\left(1-\frac{1}{999}\right)\left(1-\frac{1}{1000}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\cdot\frac{999}{1000}\)

\(=\frac{1\cdot2\cdot3\cdot....\cdot998\cdot999}{2\cdot3\cdot4\cdot....\cdot999\cdot1000}=\frac{1}{1000}\)

Vậy \(P=\frac{1}{1000}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)

\(\frac{1.51}{50.2}=\frac{51}{100}\)

a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)

\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)

Vậy \(A=\frac{1}{999}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{1000}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)

\(=\frac{1.2.3.4....999}{2.3.4....1000}\)

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}\)

\(=\frac{3.8.15....2499}{4.9.16....2500}\)

\(=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4....50.50}\)

\(=\frac{\left(1.2.3.4.5...49\right)\left(3.4.5....51\right)}{\left(2.3.4....50\right).\left(2.3.4...50\right)}\)

\(=\frac{1.51}{50.2}=\frac{51}{100}\)

\(=\frac{1}{1000}\)

= (1/1 -1/2) + (1/2-1/3) + (1/3x1/4)+...+(1/999- 1/1000)

= 1/1- 1/1000

= ...[bn tự tính nhé]

k mk nha, nếu đúng

\(\text{Đề bạn bị sai thì phải ????? Đề đúng phải là }:\)

      \(\frac{1}{1\text{ x }2}+\frac{1}{2\text{ x }3}+\frac{1}{3\text{ x }4}+...+\frac{1}{999\text{ x }1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

1/1x2+1/2x3+1/3x4+...

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

= 1-1/4

=3/4

K nhé

bạn trả lời đúng rồi nhưng giúp mình phần sau nữa nhé !

ARIGATO

1/1x2+1/2x3+...+1/999x1000+1

= 1+ (1/1-1/2+1/2-1/3+...+1/999-1/1000)

= 1 + ( 1-1/1000)

= 1 + 999/1000

= 1999/1000

K mk nhé bn

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{999}-\frac{1}{1000}+1\)

=\(\frac{1}{1}-\frac{1}{1000}+1\)

=\(\frac{1999}{1000}\)

     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

= \(1-\frac{1}{1000}+1\)

= \(\frac{999}{1000}+1\)

=\(\frac{1999}{1000}\)

                                                     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

                                                   = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

                                                   = \(\frac{1}{1}-\frac{1}{1000}+1\)

                                                   = \(\frac{999}{1000}+1\)

                                                   = \(\frac{999}{1000}+\frac{1000}{1000}\) 

                                                   = \(\frac{1999}{1000}\)