dễ

ta tách ra xog dùng phương pháp loại trừ đó

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)

\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)

Câu hỏi là:

HÃY TÍNH NHANH TỔNG SAU

\(\frac{1}{1.3.7}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{3.7}\right)\)

\(\frac{1}{3.7.9}=\frac{1}{6}\left(\frac{1}{3.7}-\frac{1}{7.9}\right)\)

....

\(\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{13.15}-\frac{1}{15.19}\right)\)

Cộng các vế với nhau ta được

\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+...+\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{37}{3.15.19}\)

h chỉ còn là nhx kỉ niệm :(

a) A = 1/3 - 1/7 + 1/7 - 1/11 +......+1/107 - 1/111

A = 1/3 - 1/111

A = ..............Bạn tự tính nhé!

b) B = 2.(3/15.18 + 3/18.21 +........+3/87.90)

B = 2.(1/15 - 1/18 + 1/18 - 1/21 +........+1/87 - 1/90)

B = 2.(1/15 - 1/90)

B = 2.5/90

B =......Tự tính nhé!

C ; D làm tương tự nhé!

yêu cầu là gì vậy

a)

A=\(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{107.111}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-.....+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{108}{333}\)

\(\frac{15}{16}\)nha bạn

úm ba la xin tích

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)

\(=1.\frac{15}{16}=\frac{15}{16}\)

3/1.4+3/4.7+3/7.10+3/10.13+3/13.16=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16=1-1/16=15/15

A=\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)

A=\(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)

A=\(2.\left(1-\frac{1}{13}\right)\)

A=\(2.\frac{12}{13}=\frac{24}{13}\)

A=2(2/1.3+2/3.5+2/5.7+...+2/11.13)

A=2(1/1-1/3+1/3-1/5+1/5-1/7+...+1/11-1/13)

A=2(1/1-1/13)=2.12/13=24/13

Ta có:

\(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{11.13}\)

\(A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)\)

\(A=2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(A=2\left(\frac{1}{1}-\frac{1}{13}\right)\)

\(A=2.\frac{12}{13}\)

\(A=\frac{24}{13}\)

Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)