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 ~ 

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 !

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

\(A=100\cdot\left(1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{9899}{9900}\right)\\ =100\cdot\left(1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{9900}\right)\\ =100\cdot\left[\left(1+1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\dfrac{49}{100}\right]\\ =100\cdot\dfrac{9851}{100}\\ =9851\)

Có: \(A=\frac{1}{2}+\frac{5}{6}+...+\frac{9899}{9900}\)

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

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

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

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

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

\(\Rightarrow A< 99\)

Mấy câu như này tách ra kiểu gì?

\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)

\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{71}{72}+\frac{89}{90}=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=8-\frac{2}{5}=\frac{38}{5}\)