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\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-....-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\frac{1}{99}+1=\frac{100}{99}\)
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{99}+\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{99}-1\right)\)
\(=-\frac{98}{99}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)
\(\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)
\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)
Đúq nhaaa
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}=-\frac{49}{50}\)
=> 50C = \(50.\left(-\frac{49}{50}\right)=-49\)
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 :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
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
= (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
=1
Đặt \(A=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(A=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-\cdots-\left(\frac12-\frac13\right)-\left(\frac11-\frac12\right)\)
\(A=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-\cdots-\frac12+\frac13-1+\frac12\)
\(A=\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}-\frac{1}{97}\right)+\left(\frac{1}{96}-\frac{1}{96}\right)+\cdots+\left(\frac12-\frac12\right)-1\)
\(A=0+0+0+0+\cdots+0+\left(-1\right)\)
\(A=-1\)
Vậy A = -1