\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{3\cdot1}\)

\(B=-\left(\dfrac{1}{3\cdot1}+\dfrac{1}{5\cdot3}+...+\dfrac{1}{97\cdot99}\right)\)

\(2B=-\left(\dfrac{2}{3\cdot1}+\dfrac{2}{5\cdot3}+...+\dfrac{2}{99\cdot97}\right)\)

\(2B=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(2B=-\left(1-\dfrac{1}{99}\right)\)

\(2B=-\dfrac{98}{99}\)

\(B=-\dfrac{98}{198}\)

`#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`

a: =11/7(-3/7+4/11-4/7+7/11)=0

b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)

\(T=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)

\(T=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

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

`#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:

\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)

\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)

\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)

\(99B=-\dfrac{4751}{97}\)

\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)

K bit có đúng k nhưng cứ nói thử k/q =-6148/15345

\(\dfrac{-4751}{9603}\)

\(\dfrac{1}{99}-\)1

Lời giải:
a)

\(=\left(\frac{-3}{7}+\frac{4}{11}+\frac{-4}{7}+\frac{7}{11}\right):\frac{7}{11}=\left(\frac{-3-4}{7}+\frac{4+7}{11}\right):\frac{7}{11}=(-1+1):\frac{7}{11}=0\)

b)

Đặt biểu thức là $A$

\(-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}-\frac{2}{97.99}\)

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{97-95}{95.97}-\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}-\frac{2}{97.99}\)

\(=1-\frac{1}{97}-\frac{2}{97.99}=\frac{96.99-2}{97.99}\)

\(\Rightarrow A=\frac{1-48.99}{97.99}\)

a,

Đặt A = \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)

\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{95}+...+\dfrac{1}{3}-1\right)\)

2A= \(2.\left(\dfrac{1}{99}-1\right)\)

\(\Rightarrow\) A = \(\dfrac{1}{99}-1=\dfrac{-98}{99}\)

b, \(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)

= \(\dfrac{3.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}{5.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)

= \(\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)

Chúc bn hc tốt <3

Ta có:

\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

\(=\dfrac{1}{99.97}=\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=\dfrac{1}{9603}-\dfrac{48}{97}\)

\(=\dfrac{-4751}{9603}\)

Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=\dfrac{-4751}{9603}\)