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

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

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

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

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

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

\(=1-\frac{1}{2018}=\frac{2017}{2018}\)

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

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

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

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

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

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

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

x= -2018/2017 

Bài làm:

Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)

\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)

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

\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)

\(\Leftrightarrow x.\frac{2017}{2018}=-1\)

\(\Rightarrow x=-\frac{2018}{2017}\)

\(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{2017\cdot2018}=-1\)

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

=> \(\frac{x}{1}-\frac{x}{2018}=-1\)

=> \(\frac{2018x-x}{2018}=-1\)

=> \(\frac{2017x}{2018}=-1\)

=> 2017x = -2018

=> x = -2018/2017

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

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

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

\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)

Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)

               \(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)

      .............................................

           \(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)

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

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

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

Chúc bạn học tốt nhớ k mình nhá

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

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

\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)

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

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

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

1/1x2+1/2x3+1/3x4+...+1/999x1000

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

=1-1/1000

=1000/1000-1/1000

=999/1000

999 / 1000

Ta có công thức tổng quát là:

\(\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)

Thay vào sẽ là:

\(\frac{2017.2018.\left(2.2017+1\right)}{6}=2737280785\)

= 2737280785

chỉ nhiu đó thui hả bạn

1999/1000

tớ gặp bài này rồi, k nhé

đúng thế còn cách làm tớ biết rồi!

noi cach lam di

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)

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

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

=> A = \(1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

2 nha bạn