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

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

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

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

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

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

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

Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

Gọi B = 1x2 + 2 x 3 + 3 x 4 + ... + 2016 x2017

    3B = 3 x ( 1x2 + 2x3 + 3x4 + ... + 2016x2017)

         = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2016x2017x3 )

         = 1x2x3 + 2x3x( 4-1) + 3x4x( 5 -2 ) + ... + 2016x2017x( 2018 - 2015)

         = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2016x2017x2018 - 2015x2016x2017

         = 2016 x2017 x2018

      B = 672 x2017 x2018

Mà A = \(\frac{672x2017x2018}{2017x2018}\)

         =  672

Vậy A = 672

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Đặt \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+....+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=\(1-\frac{1}{9}\)

=\(\frac{8}{9}\)

OK XONG NHỚ CHO MIK NHA

\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)

=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)

=1-\(\frac{1}{9}\)

=\(\frac{8}{9}\)

\(\frac{1}{1\times2}+........+\frac{1}{8\times9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

1/1x2+1/2x3+...+1/49x50

=1-1/2+1/2-1/3+.....+1/49-1/50

=1-1/50(1)

Ta co   1(2)

So sanh (1) voi (2) ta thay 1-1/50<1

=>1/1x2+...+1/49x50<1

(Phuong phap khu)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)

Vậy \(\frac{49}{50}<1\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}=\frac{15}{16}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

Cảm ơn nhé

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

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

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

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

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

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

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