Tính : \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}=...\)
(Kết quả là phân số tối giản nhé!)
\(T=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)
\(T=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
Đặt \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)=\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{48}{97}\)
Thay \(A\) vào \(T\) ta có:\(T=\dfrac{1}{99\cdot97}-\dfrac{48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)
Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)
Đặt \(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2B=1-\dfrac{1}{97}\)
\(2B=\dfrac{96}{97}\)
\(B=\dfrac{96}{97}:2\)
\(B=\dfrac{48}{97}\)
\(\Rightarrow A=\dfrac{1}{99.97}-\dfrac{48}{97}\)
\(A=\dfrac{1}{99.97}-\dfrac{48.99}{97.99}\)
\(A=\dfrac{1-48.99}{99.97}\)
\(A=-\dfrac{4751}{9603}\)
Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=-\dfrac{4751}{9603}\)
mk tick nhầm, 2 ng sai hoàn toàn rồi, chỉ có 5 phân số cộng lại thôi
Tính:1/99.97-1/97.95-1/95.93-1/5.3-1/3.1=?
(Nhập kết quả dưới dạng phân số tối giản)
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`
Tính \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-\frac{1}{5.3}-\frac{1}{3.1}\)
( Nhập kết quả dưới dạng phân số tối giản )
a)\(\frac{1}{99.97}\)−\(\frac{1}{97.95}\)−\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)−\(\frac{1}{3.1}\)
=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(\(\frac{1}{95}\)−\(\frac{1}{97}\)+\(\frac{1}{93}\)−\(\frac{1}{95}\)+…+\(\frac{1}{3}\)−\(\frac{1}{5}\)+1−\(\frac{1}{3}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)
=\(\frac{-4751}{9603}\)
Tính : 1/99.97 - 1/ 97.95 - 1/ 95.93 - 1/ 5.3 - 1/ 3.1
( Nhập kết quả dưới dạng phân số tối giản )
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
=\(\dfrac{1}{99.97}-\)(\(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\))
=\(\dfrac{1}{99.97}-\)\(\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)
=\(\dfrac{-4751}{9603}\)
Tính \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
Mình sửa lại chút.
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{99.97}-\left\{\dfrac{1}{97.95}+\dfrac{1}{95.93}\right\}-\left\{\dfrac{1}{5.3}+\dfrac{1}{3.1}\right\}\)
\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\left\{\dfrac{1}{97}+\dfrac{1}{93}\right\}-\dfrac{1}{3}.\left\{\dfrac{1}{5}+\dfrac{1}{1}\right\}\)
\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\dfrac{190}{97.93}-\dfrac{1}{3}.\dfrac{6}{5}\)
\(=\dfrac{1}{99.97}-\dfrac{2}{97.93}-\dfrac{6}{15}\)
\(=\dfrac{1}{97}.\left\{\dfrac{1}{99}-\dfrac{2}{93}\right\}-\dfrac{2}{5}\)
\(=\dfrac{-35}{297693}-\dfrac{2}{5}\)
\(=\dfrac{-175-595386}{1488465}\)
\(=\dfrac{-595561}{1488465}\)
Tách ra và rút gọn là xong bạn nhé !!
Thực hiện phép tính:
a) \(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(\dfrac{-4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
b)\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)
1.tính \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
2. tìm x. biết : \(|x+\dfrac{1}{2}|+|x+\dfrac{1}{6}|+|x+\dfrac{1}{12}|+|x+\dfrac{1}{20}|+...+|x+\dfrac{1}{110}|=11x\)
3. tính gtri b thức C= \(2x^5-5y^3+2015\) tại x, y thỏa mãn: \(|x-1|+\left(y+2\right)^{20}=0\)
tính B=\(\dfrac{1}{99.97}\)-\(\dfrac{1}{97.95}\)-...-\(\dfrac{1}{5.3}\)-\(\dfrac{1}{3.1}\)
`#3107.101107`
\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{97\cdot99}\right)-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{9603}-\dfrac{1}{2}\cdot\dfrac{96}{97}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{9603}-\dfrac{96}{97}\right)\\ =\dfrac{1}{2}\cdot\left(-\dfrac{9502}{9603}\right)\\ =-\dfrac{4751}{9603}\)
Vậy, `B = -4751/9603.`
\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)
Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)
\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)
\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5} +...+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)
\(2C=\dfrac{96}{97}\)
\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)
Thay C vào ta được:
\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)
\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)
\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)
\(99B=-\dfrac{4751}{97}\)
\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)
