Ta có : 

\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)

\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)

\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)

\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)

\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)

\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)

\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

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

Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra : 

\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\frac{A}{100}=98-\frac{49}{100}\)

\(\frac{A}{100}=\frac{9751}{100}\)

\(A=\frac{9751}{100}.100\)

\(A=9751\)

Vậy \(A=9751\)

Chúc bạn học tốt ~ 

ta có :   1/2+5/6+...+9899/9900=1/1.2+1/2.3+...+9899/99.100          =1/1-1/2+1/2-1/3+...+1/99-1/100                                                                    Tiếp theo , bạn nhìn có các phân số nào chia hết cho nhau thì gạch chúng đi....                                                                                    VD:1/2 và 1/2 (bạn nhìn ở phía trên , là 2 số đứng gần nhau đó , thấy chưa)                                                                                            - Chúng ta gạch 2 phân số đó đi , cứ tiếp tục gạch các ps tương tự:1/3;1/3;................. cho đến 1/99.                                           Ta thấy 1/1 và 1/100 còn thừa ,không thể gạch cho số nào nên ta có:     1/1-1.100=99/100                                                                                VẬY TỔNG ĐÓ LÀ 99/100

A= ​​\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2005.2006}\)= \(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2005}\)-\(\frac{1}{2006}\)=

= 1-\(\frac{1}{2006}\)= \(\frac{2005}{2006}\)

a)Ta có:\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(\Rightarrow A=\frac{2005}{2006}\)

b)Ta có:\(\frac{2005}{2006}-1=-\frac{1}{2006}\)

        Vì \(\frac{2005}{2006}\) trừ 1 được một số âm thì chứng tỏ \(\frac{2005}{2006}\)<1

Vậy A<1

Ta co " A = 1/1,2 + 1/2,3 + ... + 1/2005,2006

>=< :  A = 1 - 1/2 + 1/2 -1/3 + ... + 1/2005 - 1/2006

<+=> :" A = 2005/2006

Ta co : 2005 /2006 - 1 = 1/2006

= 2005/2006 trí 1 một số âm thì chứng tỏ : 2005/2006 < 1

+<+> : vẬY a < 1

Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)

\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)

\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)

Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)

\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)

\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)

1/

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)

\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)

\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)

2/ 

\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)

\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

câu 1 bảo binfhd dúng nhé

Hơi nhầm nè , để tôi sửa lại đề \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\)

\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)

\(A=1+1+1+...+1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-....-\frac{1}{9900}\)

\(A=98-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\right)\)

\(A=98-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(A=98-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=98-\left(\frac{1}{2}-\frac{1}{100}\right)=98-\frac{49}{100}=\frac{9751}{100}\)

Vậy.............

 \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9989}{9900}\)

\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)

               có 50 số 1

\(A=50-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

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

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

Thay B vào A ta được:

\(A=50-\frac{49}{100}=\frac{5000}{100}-\frac{49}{100}=\frac{4951}{100}\)

cảm ơn 2 bạn nhiều nha !

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{9899}{9900}\)

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+.....+\left(1-\frac{1}{9900}\right)\)

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

\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

\(=99-\left(1-\frac{1}{100}\right)=99-1+\frac{1}{100}=98+\frac{1}{100}=\frac{9801}{100}\)

cảm ơn nha

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

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

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

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

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

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\) 

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

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

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

    \(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

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

Vậy \(B=\frac{99}{100}\)