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

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

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

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

b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)\)

\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2.\frac{2018}{2019}\)

\(=\frac{4036}{2019}\)

Phần c tương tự nha

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

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

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

câu a) mik sửa đề một tí ko biết có đúng ko

câu b , c tương tự nhưng cần lấy tử ra chung 

a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)

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

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

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

b)nhóm 2 ra ngoài rồi làm như câu a

c)nhóm 4 ra rồi làm như câu a

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

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

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

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

1999/1000

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

= 1 - 1/1000 + 1 

= 1000/1001

Tk hộ mình nhé 

1 

Mà dấu . đó là gì

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

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

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

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

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

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

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

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

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

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

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

\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)

\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)

\(\Leftrightarrow x\cdot\frac{5}{3}=15\)

\(\Leftrightarrow x=15:\frac{5}{3}\)

\(\Leftrightarrow x=15\cdot\frac{3}{5}\)

\(\Leftrightarrow x=9.\)

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

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

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

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

b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)

\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)

\(\Rightarrow x.\frac{5}{3}=14+1=15\)

\(\Rightarrow x=15:\frac{5}{3}=9\)

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

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

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

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

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

b) \(\left(x\cdot\frac{5}{3}-1\right):9=3\frac{1}{2}:2,25\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{7}{2}:\frac{9}{4}\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{14}{9}\)

\(x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(x\cdot\frac{5}{3}-1=14\)

\(x\cdot\frac{5}{3}=14+1\)

\(x\cdot\frac{5}{3}=15\)

\(x=15:\frac{5}{3}\)

\(x=9\)

Vậy \(x=9\)

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

#)Giải :

Đặt \(A=4-\frac{2}{1.2}-\frac{2}{2.3}-\frac{2}{3.4}-...-\frac{2}{99.100}\)

\(A=4-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)

\(A=4-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

\(A=4-2\left(1-\frac{1}{100}\right)\)

\(A=4-2\times\frac{99}{100}\)

\(A=4-\frac{99}{50}\)

\(A=\frac{101}{50}\)

2y= 2/ 1.2.3 + 2/2.3.4 + 2/3.4.5 +.... +2/998.999.1000

2y=1/1.2 - 1/2.3 +1/2.3 - 1/3.4 + 1/3.4 -1/4.5 +....+ 1/998.999 - 1/ 999.1000

2y=1/2 - 1/ 999.1000

2y = 499500-1 /  999.1000

2y=499499 / 999.1000

y=499499 / 1998000

Ủng hộ mk nha

bai toan nay khó quá

Ta có : \(2Y=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{998.999.1000}\)

\(\Rightarrow2Y=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{998.999}-\frac{1}{999.1000}\)

\(\Rightarrow2Y=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{1}{2}-\frac{1}{999000}=\frac{499499}{999000}\)

\(\Rightarrow Y=\frac{499499}{999000}\div2\)

\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)

x=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5

x=1-1/5

x=4/5