\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}\)

\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

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

\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)

\(\Rightarrow C=\frac{1}{50}-1\)

\(\Rightarrow50C=\left(\frac{1}{50}-1\right)50\)

\(\Rightarrow50C=1-50\)

\(\Rightarrow50C=-49\)

Vậy \(50C=-49\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

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

\(=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)

=> 50C = \(50.\left(-\frac{49}{50}\right)=-49\)

-49 đúng thì cho một lần ****

\(C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\cdot\cdot\cdot+\frac{1}{2\cdot1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

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

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

\(C=\frac{-98}{100}\)

\(Khi\) đó \(50C=\frac{-98}{100}\cdot50\)

\(50C=-49\)

tick cho mình nha

C= 1/100-(1/100.99+.....+1/3.2+1/2.1)

Ta lần lượt có :

   1/2.1=1-1/2

   1/2.3=1/2-1/3

     ................

nên C= 1/100-(1-1/100)=1/100-99/100 =-49/50

nên 50C= -49

C = \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

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

\(=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)

\(\Rightarrow50C=50.\left(-\frac{49}{50}\right)=-49\)

mk moi giai violympic =50 sao đó

mk nghi z

mk thi co 10 cau mk cung giai cau do nhung co dc 90 diem o

ma mk nghi chac cau do dung

Đặt A=  \(-\frac{1}{100.99}-\frac{1}{99.98}-....-\frac{1}{2.1}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

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

C = \(\frac{1}{100}+\left(-\frac{99}{100}\right)=-\frac{49}{50}\)

50C = \(-\frac{49}{50}.50=-49\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

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

\(C=\frac{1}{100}-1+\frac{1}{100}\)

\(C=\frac{-49}{50}\)

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)

C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)

C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)

C = 1/100 - (1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

\(c=\frac{1}{100}-\frac{1}{100.98}\frac{1}{99.98}\frac{1}{98.97}-......-\frac{1}{3.2}-\frac{1}{2.1}\)=\(\frac{1}{100}-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right]\) =\(\frac{1}{100}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\right]\)=\(\frac{1}{100}-\left[1-\frac{1}{100}\right]=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{49}{50}\)

hinh nhu la 0

C=\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}\)

Suy ra 50C=-196

C=-49/50

=>50C=-49

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100}+\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}-...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)

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

\(=-\frac{49}{50}\)

\(\Rightarrow50C=50.\frac{-49}{50}=-49\)

Ta có: \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)

\(\Rightarrow C=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(\Rightarrow C=\frac{1}{100}+\frac{1}{100}-1\)

\(\Rightarrow C=\frac{2}{100}-\frac{100}{100}\)

\(\Rightarrow C=-\frac{88}{100}=-\frac{22}{25}\)

Vậy \(C=-\frac{22}{25}\)

Chuk bạn hok tốt!