A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

Câu 1 :

1/n - 1/n + a = a + n/a ( a + n ) = a + n - a/a ( n + a ) = n/a ( a + n )

Câu 2 :

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

= 1/1 - 1/100 = 99/100

\(B=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{100.103}+\dfrac{4}{103.106}\)

\(B=4\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}+\dfrac{1}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}+\dfrac{3}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}.\dfrac{103}{318}\)

\(B=\dfrac{412}{954}\)

a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)

Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)

b,

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

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

\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)

\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)

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

\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)

\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)

\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)

\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)

\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)

\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)

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

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

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

\(=\dfrac{99}{100}\)

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy \(S=\dfrac{99}{100}\)

Chúc bạn học tốt!!!

Ta có : P = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.11}+...+\dfrac{5}{2017.2020}\)

\(\Rightarrow P=5\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{2017.2020}\right)\)

\(\Rightarrow\) \(P=5.\dfrac{3}{3}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{2017.2020}\right)\)

\(\Rightarrow P=5.\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\right)\)

\(\Rightarrow P=5.\dfrac{1}{3}\left(1-\dfrac{1}{2020}\right)\)

Lấy máy tình CASIO bấm ra kết quả nha!

Ai có nick OLM nhớ kb với mik nha . https://olm.vn/thanhvien/ngannga123!

P=\(\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{2017.2020}\)

P=\(1-\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\)

P=1-\(\dfrac{1}{2020}\)

\(\Rightarrow P=\dfrac{2019}{2020}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=5\left(1-\frac{1}{103}\right)\)

\(3B=5.\frac{102}{103}\)

\(3B=\frac{510}{103}\)

\(\Rightarrow B=\frac{170}{103}\)

Ta có:

B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)

B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)

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

B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)

B=\(\frac{5}{3}.\frac{102}{103}\)

B=\(\frac{170}{103}\)

Vậy B=\(\frac{170}{103}\)

nhớ k