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

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

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

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

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

d=1/1.2+5/2.3+11/3.4+...+9899/99.100

=>d=1-1/2+1/2-1/3+...+1/99-1/100

=>d=1-1/100

=>d=99/100

Vậy d=99/100

99/100

mk lam dc ma ko ranh ghi

ai giup minh voi

1/2 + 5/6 + 11/12 + 19/20 + 29/30 +. . . 9701 + / 9702 + 9899/9900 = 1/2 + (1-1 / 6) + (1-1 / 12) + (1-1 / 20) + (1-1 / 30) + ...... + (1 -1/9702) + (1-1 / 9900) = 1/2 + [1 - (1 / 2-1 / 3)] + [1 - (1 / 3-1 / 4)] + [1- ( 1 / 4-1 / 5)] + [1 - (1 / 5-1 / 6)] + ...... + [1- (1 / 98-1 / 99)] + [1 - (1 / 99-1 / 100)] * 100 + 1 = 1 / 2-1 / 2 + 1 / 3-1 / 3 + 1 / 4-1 / 4 + 1 / 5-1 / 5 + 1 / 6-1 / 6 + ... ... 1 / 98-1 / 98 + 1 / 99-1 / 99 + 1/100 + 1 = 100/100 = 100 và 1/100

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

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

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

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

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

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

cảm ơn nha

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

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

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

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

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

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

\(\Rightarrow A< 99\)

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

1/2 + 5/6 + 11/12 + 19/20 + 29/30 +. . . 9701 + / 9702 + 9899/9900 = 1/2 + (1-1 / 6) + (1-1 / 12) + (1-1 / 20) + (1-1 / 30) + ...... + (1 -1/9702) + (1-1 / 9900) = 1/2 + [1 - (1 / 2-1 / 3)] + [1 - (1 / 3-1 / 4)] + [1- ( 1 / 4-1 / 5)] + [1 - (1 / 5-1 / 6)] + ...... + [1- (1 / 98-1 / 99)] + [1 - (1 / 99-1 / 100)] * 100 + 1 = 1 / 2-1 / 2 + 1 / 3-1 / 3 + 1 / 4-1 / 4 + 1 / 5-1 / 5 + 1 / 6-1 / 6 + ... ... 1 / 98-1 / 98 + 1 / 99-1 / 99 + 1/100 + 1 = 100/100 = 100 và 1/100

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

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 ~