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

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

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

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

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

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

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

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

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{9x10}\)

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

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

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

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

1/2x3+1/3x4+......+1/38x39+1/39x40

=1/2-1/3+1/3-1/4+.....+1/39-1/40

=1/2-1/40

=19/40

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{9}{20}\)

\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ ... + \(\frac{1}{18x19}\)+ \(\frac{1}{19x20}\)

= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)

= \(\frac{1}{2}\)- \(\frac{1}{20}\)

= \(\frac{18}{40}\)= \(\frac{9}{20}\)

=1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20         K MIK NHA MOI NGUOI 

=1/2-1/20

=10/20-1/20

=9/20

= 9/10

k nha

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

b) \(29\times87-29\times23+64\times71=29\times\left(87-23\right)+64\times71\)

                                            \(=29\times64+64\times71=64\times\left(29+71\right)=64\times100=6400\)

c) \(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{19\times20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{9}{20}\)

B = 29 x  ( 87 - 23 ) + 64 x 71 

B = 29 x  64            + 64 x 71 

B = 64 x ( 29 + 71 )

B = 64 x 100

B = 6400

D = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)\(\frac{1}{20}\)

D = \(\frac{1}{2}-\frac{1}{20}\)

D = \(\frac{9}{20}\)

B = 29 x 87 - 29 x 23 + 64 x 71

    = 29 x (87 - 23)  + 64 x 71

   = 29 x 64 + 64 x 71

  = (29 + 71) x 64

  = 100 x 64 = 6400

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{120x121}+\frac{1}{121x122}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{120}-\frac{1}{121}+\frac{1}{121}-\frac{1}{122}\)

Ta thấy nếu + 1/3 và đồng thời - 1/3 thì đc số ban đầu nên áp dụng quy luật "trái dấu ta gạch"

\(=\)\(\frac{1}{2}-\frac{1}{122}\)

\(=\frac{30}{61}\)

P/s: Cái dòng chữ Ta thấy... bn đừng viết dzô nha, cái đó mk giải thích thêm thui cho bn hỉu

Mk lm mãi mới đc đó. Bài này ko khó cũng ko dễ chỉ thuộc loại bt thui nha 

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{120x121}+\frac{1}{121x122}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{120}-\frac{1}{121}+\frac{1}{121}-\frac{1}{122}\)

\(=\frac{1}{2}-\frac{1}{122}\)

\(=\frac{61}{122}-\frac{1}{122}\)

\(=\frac{60}{122}=\frac{30}{61}\)

Ủng hộ mk nha ^_-

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{121.122}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{121}-\frac{1}{122}\)

\(=\frac{1}{2}-\frac{1}{122}\)

\(=\frac{30}{61}\)

\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)

\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)

\(\Rightarrow100n-100=98n+98\)

\(\Rightarrow2n=198\)

=> n = 99

Vậy n =  99

\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

         \(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)

          \(\frac{1}{n+1}\)=\(\frac{1}{100}\)

=> n+1=100

        n=100-1

       n=99

\(\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{nx\left(n+1\right)}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{2}-\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{100}\)

n + 1 = 100

n = 100 - 1

n = 99

Bn k cho mik nhé!