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

\(A=1-\frac{1}{1000}=\frac{999}{1000}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)

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

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

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

1: \(A=4\dfrac{7}{1000}\cdot\dfrac{1}{999}-1\dfrac{1}{500}\cdot\dfrac{4}{999}+\dfrac{1001}{999\cdot1000}\)

Đặt 1/1000=a; 1/999=b

\(A=\left(4+7a\right)\cdot b-\left(1+2a\right)\cdot4b+b\cdot\dfrac{1001}{1000}\)

\(=4b+7ab-4b-8ab+b\cdot\left(1+a\right)\)

=-ab+b+b+ba=2b=2/999

2: Đặt 1/4587=a;1/3897=b

\(B=a\cdot\left(7+b\right)-\left(3+1-a\right)\cdot2b-7a-3ab\)

=7a+ab-8a+2ab-7a-3ab

=-8a=-8/4587

\(a,\frac{1}{999\cdot1000}-\frac{1}{998\cdot999}-\frac{1}{997\cdot998}-...-\frac{1}{2\cdot1}\)

\(=\frac{1}{999\cdot1000}-\left[\frac{1}{2\cdot1}+\frac{1}{2\cdot3}+...+\frac{1}{997\cdot998}+\frac{1}{998\cdot999}\right]\)

\(=\frac{1}{999\cdot1000}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{998}-\frac{1}{999}\right]\)

\(=\frac{1}{999\cdot1000}-\left[1-\frac{1}{999}\right]=\frac{1}{999\cdot1000}-\frac{998}{999}=...\)

Tính nốt , không chắc :v

a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)

b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)

Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)

\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)

Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)

bn lm sai rồi

AI KẾT BN KO!

TIỆN THỂ TK MÌNH LUÔN NHA!

KONOSUBA!!!

A=\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{999.1000}\)\

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

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

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

a) \(\left(\frac34+\frac{-7}{2}\right).\left(\frac{2}{11}+\frac{12}{22}\right)\)

\(=\left(\frac34+\frac{-14}{2}\right).\left(\frac{2}{11}+\frac{6}{11}\right)\)

\(=-\frac{11}{4}.\frac{8}{11}=-2\)

b) \(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\ldots\frac{999^2}{999.1000}=\frac12.\frac23.\frac34.\ldots\frac{999}{1000}\)

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

c) ta phân tách 30=5.6

42=6.7

56=7.8

\(\vert\)

132=11.12

thay vào biểu thức:

= \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\cdots\frac{1}{11.12}\)

ta có công thức đã dc học: \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

=> \(\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\frac18-\frac19+...+\frac{1}{11}-\frac{1}{12}\)

= \(\frac15-\frac{1}{12}=\frac{7}{60}\)