a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)

b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)

b) A=1/2.3+1/3.4+....+1/99.100

=> A=1/2-1/3+1/3-1/4+....+1/99-1/100

=> A=1/2-1/100

=> A=50/100-1/100

=> A=49/100

49/100 

k nhe

49/100

b) 

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

\(B=\frac{1}{5}-\frac{1}{12}\)

\(B=\frac{7}{60}\)

a) Ta có: 

\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)  ;   \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{1\left(n+1\right)}\)

Vậy \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)

b)  \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{100-99}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

\(B=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-\frac{1}{11}-\frac{1}{12}\)

\(B=\frac{1}{5}-\frac{1}{12}\)

\(B=\frac{7}{60}\)

\(\frac{7}{60}\) ok

N = 1 - 2/2.3 + 1 - 2/3.4 +.....+ 1 - 2/99.100

   = 98 - 2.(1/2.3 + 1/3.4 + ...... + 1/99.100)

   = 98 - 2.(1/2-1/3+1/3-1/4+....+1/99-1/100)

   = 98 - 2.(1/2-1/100)

   = 98 - 2.49/100 = 98-49/50 < 98

Mà 49/50 < 1

=> N > 98-1 = 97

=> 97 < N < 98

Tk mk nha

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}=\frac{5}{6}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)

ui cí này e chưa học

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}\)

\(=\frac{5}{6}\)

cau a),b),c) ban dat mau chung roi khu mau ma lam la duoc ma

Tuy học lớp 6 ................. cơ mừ thấy mí bài nỳ dễ quá >.<

Ta có công thức tổng quát của số hạng trong tổng trên có dạng:

\(x_n=\frac{n\left(n+3\right)}{\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n+2-2}{n^2+3n+2}\)

\(=1-\frac{2}{n^2+3n+2}=1-\frac{2}{\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow\frac{1.4}{2.3}=1-\frac{2}{2.3}\)

\(\frac{2.5}{3.4}=1-\frac{2}{3.4}\)

\(\frac{3.6}{4.5}=1-\frac{2}{4.5}\)

....

\(\frac{98.101}{99.100}=1-\frac{2}{99.100}\)

\(\Rightarrow N=98-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(=98-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98-2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=98-1+\frac{1}{50}=97+\frac{1}{50}\)

Vậy 97 < N < 98

`1/(2.3)+1/(3.4)+......+1/(99/100)`
`=1/2-1/3+1/3-1/4+..........+1/99-1/100`
`=1/2-1/100`
`=49/100`

Đặt A= \(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{99.100}\)

      A= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\)\(\dfrac{1}{99}-\dfrac{1}{100}\)

     A=\(\dfrac{1}{2}-\dfrac{1}{100}\)

   A=\(\dfrac{49}{100}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) 

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\) 

\(=\dfrac{1}{2}-\dfrac{1}{100}\) 

\(=\dfrac{49}{100}\)