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 \(\frac{7}{222222}\))
= \(\frac{7}{22}\)

Ta có: 2/1.3 = 1/1 - 1/3

          2/3.5 = 1/3 - 1/5

\(\Rightarrow\) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

=   1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/100

=   1 - 1/100

=    99/100

tích trên sẽ = 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100

=1-1/100 =99/100

bạn nhớ rằng  k/n.(n+k) sẽ = 1/n-1/n+k

=1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

=1-1/101

=100/101

Đúng 100 phần trăm luôn

\(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{99.101.103}\)

=\(\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{99.101.103}\right)\)

=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)\)

=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{101.103}\right)\)

=\(\frac{1}{4}.\frac{10406}{31209}\)

=\(\frac{5230}{62418}\)

tui chịu thôi

Ta có: 1/1.3.5 = (1/1.3 - 1/3.5).1/4

          1/3.5.7 = (1/3.5 - 1/5.7).1/4

\(\Rightarrow\) 1/1.3.5 + 1/3.5.7 + 1/5.7.9 + ... + 1/99.101.103 = 1/4.(1/1.3 - 1/3.5 + 1/3.5 - 1/5.7 + ... + 1/99.101 - 1/101.103)

                                                                           = 1/4.(1/3 - 1/10403)

                                                                           = 2600/31209

Tớ nghĩ vậy, nếu đúng thì cho mk biết nha

a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)

   \(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)

\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)

b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)

\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)

\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)

\(\Rightarrow x=10\cdot\)

Bai nay mk vua giai xong nen ko muon giai nua

mk chi neu cach giai thoi

dau tien tinh 1/2P ra quy luat o mau la 2.6 ; 6.10 ; 10.14 ; ...; 198.202 roi nhan voi 4 la ra

Ket qua la 25/101

kết quả là \(\frac{25}{101}\)

\(P=\frac{1}{2.3}+\frac{1}{6.5}+............+\frac{1}{198.101}\)

\(P=\frac{1}{6}+\frac{1}{30}+.........+\frac{1}{19998}\)

\(\frac{1}{2}P=\frac{1}{12}+\frac{1}{60}+...........+\frac{1}{39996}\)

\(\frac{1}{2}P=\frac{1}{2.6}+\frac{1}{6.10}+.........+\frac{1}{198.202}\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+..........+\frac{4}{198.202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+..........+\frac{1}{198}-\frac{1}{202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(\frac{1}{2}P=\frac{1}{4}.\frac{50}{101}\)

\(\frac{1}{2}P=\frac{25}{202}\)

\(P=\frac{25}{202}:\frac{1}{2}\)

\(P=\frac{25}{101}\)

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)

    \(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+\frac{2}{14.18}+...+\frac{2}{198.202}\)

     \(=\frac{1}{2}.\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+\frac{4}{14.18}+...+\frac{4}{198.202}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)

    \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

    \(=\frac{1}{2}.\frac{50}{101}=\frac{25}{101}\)

Ai đúng mình k

where ?

khoan đã bạn chép nhầm đề rồi thì phải số 1 kia không có dấu gì à?

Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

\(\Rightarrow A=\frac{98}{303}\div2=\frac{49}{303}\)

#)Giải :

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

= \(\frac{1}{5}-\frac{1}{25}\)

= \(\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

= \(1-\frac{1}{101}\)

= \(\frac{100}{101}\)

c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)

= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)

= \(5\frac{2}{7}\)

= \(\frac{37}{7}\)

a)\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

đề có sai ko nhỉ ???