1-1/1000

999/1000

Ta có: 1/1.2 + 1/2.3 +1/3.4 +......+1/998.999 + 1/999. 1000

         = 1/2 + 1/6 + 1/12 + .... + 1/997002 + 1/999000

 lại có : 1/2 = 1-1/2

            1/6  = 1/2 -1/3

             1/12 = 1/3 - 1/4 

             ...

           1/997002 = 1/998 - 1/999

           1/999000 = 1/999 - 1000

=>1/1.2 + 1/2.3 +1/3.4 +......+1/998.999 + 1/999. 1000

  = 1-1/2 + 1/2 - 1/3 + 1/3 -1/4 +....+ 1/998 - 1/999 + 1/999 - 1/1000

  = 1-1/1000

  = 999/1000

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{999^2}{999.1000}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{999.999}{999.1000}\)

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

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

A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)

A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)

B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)

B-A=1/1008

sách 6,7,8 có 2 bài này nè. mk k bt ghi ps nên mk ko gửi đc sorry nha. Hhh

a)\(A=\frac{10^{2014}+2016}{10^{2015}+2016}=>10A=\frac{10^{2015}+20160}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\left(1\right)\)

\(B=\frac{10^{2015}+2016}{10^{2016}+2016}=>10B=\frac{10^{2016}+20160}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2106}\left(2\right)\)

từ 1 zà 2 

=> 10A>10B

=>A>B

b) Đặt Biểu thức trong ngoạc là A nha

=>3A\(=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}...+\frac{3}{7.8.9.10}\)

=>3A\(=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+\frac{6-3}{3.4.5.6}+...+\frac{10-7}{7.8.9.10}\)

=>3A=\(\frac{1}{1.2.3.}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{7.8.9}-\frac{1}{8.9.10}\)

=>3A=\(\frac{1}{1.2.3}-\frac{1}{8.9.10}=>A=\frac{1}{1.2.3.3}-\frac{1}{8.9.10.3}=\frac{119}{2160}\)

=>x=\(\frac{119}{120}:\frac{119}{2160}=18\)

\(E=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2016.2018}\)

\(E=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2018-2016}{2016.2018}\)

\(2E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)

\(E=\left(\frac{1}{2}-\frac{1}{2018}\right).\frac{1}{2}\)

\(E=\frac{504}{1009}.\frac{1}{2}\)

\(E=\frac{252}{1009}\)

\(E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\)

\(E=\frac{1}{2}-\frac{1}{2018}\)

\(E=\frac{1005}{2018}\)

\(E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2116}-\frac{1}{2018}\)

\(E=\frac{1}{2}-\frac{1}{2018}\)

\(E=\frac{1005}{2018}\)

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

\(=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{1000-999}{999\cdot1000}+1\)

\(=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+...+\frac{1000}{999\cdot1000}-\frac{999}{999\cdot1000}+1\)

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

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

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

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

\(\text{Đề phải như này bạn nha : }B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)

N Lam theo đề Nguyễn Thiều Công Thành nha  :

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

\(\Rightarrow B=1-\frac{1}{1000}=\frac{999}{1000}\)

sai đề rồi bạn ơi

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

=1-1/100+1

=199/100