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a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\frac{99}{100}\)
\(A=-\frac{98}{100}=-\frac{49}{50}\)
Các bạn giúp mk với, mk sắp phải nộp rồi. Ai nhanh nhất mk k cho
\(-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..........-\frac{1}{2.1}\)
\(=-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..........+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
1/100‐1/100.99‐1/99.98‐...‐1/3.2‐1/2.1
\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}\)
\(\frac{1}{100-99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{2.1}\right)\)
\(\frac{1}{100-99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{99}-\frac{1}{100}-\frac{98}{99}\)
\(-\frac{97}{99}-\frac{1}{100}\)
\(-\frac{9799}{9900}\)
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{2\cdot1}\)
\(=\frac{1}{100\cdot99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99\cdot100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}\)
\(=\frac{-9799}{9900}\)
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
= (1/99-1/100)- (1/98-1/99)-...(1/1-1/2)
= -(1/1-1/2+1/3-1/4+...+1/99-1/100)
=-(1/1-1/100)
=-99/100
trong câu hỏi tương tự rõ hơn
C = 1/100 - 1/100 x 99 - 1/99 x 98 + 1/98 x 97 - ..- 1/3 x 2 - 1/2 x 1
C = 1/100 - ( 1/100 x 99 - 1/99 x 98 + 1/98 x 97 - ... - 1/3 x 2 - 1/2 x 1 )
C = 1/100 - ( 1/1 x 2 - 1/2 x 3 - .....- 1/97 x 98 - 1/98 x 99 - 1/99 x 100 )
C = 1/100 - ( 1 - 1/2 + 1/2 - 1/3 + .... + `1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )
C = 1/100 - ( 1 - 1/100 )
C = 1/100 - 99/100
C = 49/50
C=-(1/1.2+1/2.3+.....+1/99.100+1/100)=-(1/1-1/2+1/2-1/3+....+1/99-1/100+1/100)=-(1-1/100+1/100)=-1
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-........-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+........+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
\(\Rightarrow50C=-\frac{49}{50}.50=-49\)
tick nha
\(C=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)+\frac{1}{100}\)
\(=>C=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{100}\)
=>\(C=-\left(\frac{1}{1}-\frac{1}{100}\right)+\frac{1}{100}\)
=>\(C=-\frac{99}{100}+\frac{1}{100}=-\frac{98}{100}\)
=>50C=\(-\frac{98}{100}.50=\frac{-98}{2}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-...\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...\frac{1}{2.1}\right)\)
\(=\frac{1}{100.99}-\left(\frac{1}{99}-1\right)\)
\(=\frac{1}{9900}-\frac{-98}{99}=\frac{1}{9900}+\frac{98}{99}=\frac{1}{9900}+\frac{9800}{9900}=\frac{9800}{9900}\)