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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x=10\cdot\)

đề có sai ko nhỉ ???

đây mà là toán lớp 1 à

 toán lớp 1 à nói đi lớp mấy

toán lớp 6 nhé

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

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

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

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

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

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

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

Đặt biểu thức trên là A = 1/1.3 + 1/3.5 + 1/5.7 +...+ 1/99.101

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

2A= 2/1.3 + 2/3.5 + 2/ 5.7 +... 2/99.101

2A= 1-1/3 + 1/3 - 1/5 + 1/5-1/7 +...+ 1/99-1/101

2A= 1-1/101

2A= 100/101

A= 100/101 . 1/2

A= 50/101

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

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

\(\Rightarrow1-\frac{1}{101}=\frac{100}{101}\)

Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{99.101}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101

A = 2 - 2/101 = 200/101

B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51

B = 3-3/51(tự tính nhé)

C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31

C = 5(5-1/31)(tự tính)

D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)

2E nhân lên rồi giải giống trên

3F Rồi nhân 4/77 và rút gọn thì tính được

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

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

A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0

A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)

a) A= \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\) 

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

=\(1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)

=\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right).\frac{3}{2}\)

=\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right).\frac{3}{2}\)

= \(\left(1-\frac{1}{50}\right).\frac{3}{2}=\frac{49}{50}.\frac{3}{2}=\frac{147}{100}\)

c) \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

= \(\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right).5\)

= \(\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right).5\) 

= \(\left(1-\frac{1}{31}\right).5=\frac{30}{31}.5=\frac{150}{31}\)

Mấy bài còn lại mik đang phải nháp đã. Bạn thông cảm cho mik

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

\(A=\frac{2}{3}+\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{2}{3}+\frac{98}{303}\)

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

\(A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+.....+\frac{2}{99x101}\)

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

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

\(A=\frac{99}{100}\)

A=2/1*3+2/3*5+2/5*7+....+2/99*101

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

A=1-1/101

A=100/101