1/99-1/99.97-1/97.95-1/95.93-...+1/5.3-1/3.1
1/99.97-1/97.95-1/95.93-...-1/5.3-1/3.1 = ?
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-...-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)\(=\frac{1}{99\cdot97}-\left(\frac{1}{97\cdot95}+\frac{1}{95\cdot93}+...+\frac{1}{3\cdot1}\right)\)
\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-\frac{1}{95}+\frac{1}{95}-\frac{1}{93}+...+\frac{1}{3}-1\right)\)\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-1\right)=\frac{1}{9603}-2\cdot\left(-\frac{96}{97}\right)\)\(\frac{1}{9603}-\frac{-192}{97}\)phần còn lại tự làm
1/99.97-1/97.95-1/95.93........1/5.3 - 1/3.1
Đặt A=\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-........-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
=\(\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+......+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+.......+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\) =\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)
=\(\dfrac{-4751}{9603}\)
Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\\ \Rightarrow 2A= \dfrac{2}{99.97}-\dfrac{2}{97.95}-\dfrac{2}{95.93}-...-\dfrac{2}{5.3}-\dfrac{2}{3.1}\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97})-(\dfrac{1}{93}-\dfrac{1}{95})-...-(\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A = \dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-1+\dfrac{1}{97}\\ \Rightarrow A\)
1/(99.97)-1/(97.95)-1/(95.93)-...-1/(5.3)-1/(3.1). Tính nhanh
1/99.97-1/97.95-1/95.93...-1/5.3-1/3.1
Ta có : \(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
= \(\frac{1}{99.97}-\left(\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
=\(\frac{1}{99.97}-\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{97}+\frac{1}{93}-\frac{1}{95}+...+\frac{1}{3}-\frac{1}{5}+1-\frac{1}{3}\right)\)
= \(\frac{1}{99.97}-\frac{1}{2}.\left(1-\frac{1}{97}\right)\)
= \(\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)
= \(\frac{1}{99.97}-\frac{48}{97}=\frac{1}{99.97}-\frac{48.99}{99.97}=\frac{-4751}{9603}\)
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{93.95}+\frac{2}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)
\(=\frac{1}{9603}-\frac{48}{97}=\frac{-4751}{9603}\)
Tính: A = 1/99.97- 1/97.95- 1/95.93- ..... -1/5.3- 1/3.1
A=-(1/1.3+1/3.5+1/5.7+...+1/97.99)
A=-1/2.(2/1.3+2/3.5+2/5.7+...+2/97.99)
A=-1/2.(1-1/3+1/3-1/5+...+1/97-1/99)
A=-1/2.(1-1/99)=-1/2.98/99
A=(tự bấm máy tính nha)
lam j co tru o dang trc 1/99*97 sai tram trong
1/99.97 - 1/97.95 - 1/95.93 - ... - 1/5.3 - 1/3.1. Thực hiện phép tính
Tinh:A=1/99.97-1/97.95-1/95.93-...-1/5.3-1/3.1
Tính:
1/99.97-1/97.95-1/95.93-....-1/5.3-1/3.1
\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\cdot\dfrac{96}{97}\)
\(=\dfrac{1}{97\cdot99}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)
Tính: \(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{3\cdot1}\)
\(B=-\left(\dfrac{1}{3\cdot1}+\dfrac{1}{5\cdot3}+...+\dfrac{1}{97\cdot99}\right)\)
\(2B=-\left(\dfrac{2}{3\cdot1}+\dfrac{2}{5\cdot3}+...+\dfrac{2}{99\cdot97}\right)\)
\(2B=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(2B=-\left(1-\dfrac{1}{99}\right)\)
\(2B=-\dfrac{98}{99}\)
\(B=-\dfrac{98}{198}\)
`#3107`
`B = 1/(99*97) - 1/(97*95) - 1/(95*93) - ... - 1/(5*3) - 1/(3*1)`
`= 1/(99*97) - (1/(1*3) + 1/(3*5) + ... + 1/(95*97) )`
`= 1/2*(2/(97*99) ) - 1/2*(2/(1*3) + 2/(3*5) + ... + 2/(95*97) )`
`= 1/2*(1/97 - 1/99) - 1/2*(1 - 1/3 + 1/3 - 1/5 + ... + 1/95 - 1/97)`
`= 1/2*(1/97 - 1/99) - 1/2*(1 - 1/97)`
`= 1/2*(1/97 - 1/99 - 1 + 1/97)`
`= 1/2*(-9502/9603)`
`= -4751/9603`