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Đặt S = | 1 | + | 1 | + … + | 1 |
1 . 3 | 3 . 5 | 99 . 101 |
1 | - | 1 | = | 3 - 1 | = | 2 |
1 | 3 | 1 . 3 | 1 . 3 |
1 | = | 1 | ( | 1 | - | 1 | ) |
1 . 3 | 2 | 1 | 3 |
1 | = | 1 | ( | 1 | - | 1 | ) |
3 . 5 | 2 | 3 | 5 |
1 | = | 1 | ( | 1 | - | 1 | ) |
5 . 7 | 2 | 5 | 7 |
1 | = | 1 | ( | 1 | - | 1 | ) |
99 . 101 | 2 | 99 | 101 |
S = | 1 | ( | 1 | - | 1 | ) |
2 | 1 | 101 |
S = | 1 | 101 - 1 | |
2 | 101 |
S = | 100 |
202 |
Tổng ban đầu = | 50 |
101 |
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi biểu thức đó là A
\(A=\frac{3}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right]\)
\(A=\frac{3}{2}\left[1-\frac{1}{101}\right]\)
\(A=\frac{3}{2}.\frac{100}{101}=\frac{300}{202}=\frac{150}{101}\)
Mk nghĩ là vậy đó
Chúc bạn học tốt !
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
B = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{630}\)
B = \(1+\frac{2}{6}+\frac{2}{12}+\frac{1}{20}+...+\frac{2}{1260}\)
B = \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{35.36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{36}\right)=1+2.\frac{17}{36}\)
B = \(1+\frac{17}{18}\)
B = \(\frac{35}{18}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{99}{101}\)
\(D=\frac{1\cdot3\cdot5\cdot7\cdot...\cdot99}{3\cdot5\cdot7\cdot9\cdot...\cdot101}\)
\(D=\frac{1}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Link nè lên google search nha!
https://olm.vn/hoi-dap/question/162533.html
A = \(\frac{1}{1\cdot3}\)+ \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ ..... + \(\frac{1}{99.101}\)
= \(\frac{1}{2}\). ( \(\frac{1}{1.3}\)+ \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ ...... + \(\frac{1}{99.101}\))
= \(\frac{1}{2}\). ( 1 - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ........ + \(\frac{1}{99}\)- \(\frac{1}{101}\))
= \(\frac{1}{2}\). ( 1 - \(\frac{1}{101}\))
= \(\frac{1}{2}\). \(\frac{100}{101}\)= \(\frac{50}{101}\)
Thấy đúng thì cho mình một k nha!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt tên cho biểu thức là A
A x 2/3 = 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9+ ... + 2/99x101
Ax2/3 = 3-1/1x3 + 5-3/3x5 + 7-5/5x7 + ... + 101/99
Ax2/3= 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
Ax2/3=1-1/101
Ax2/3=100/101
A=100/101:2/3
A=150/101
tổng trên sẽ là:
3/1x3+....x101=2007
đáp số:2007
k mình nha bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=1x3+3x5+5x7+7x9+...+99x101
6A=6x(1x3+3x5+5x7+7x9+...+99x101)
6A=1x3x6+3x5x6+5x7x6+7x9x6+...+99x101x6
6A=1x3x(5+1)+3x5x(7-1)+5x7x(9-3)+7x9x(11-5)+...+99x101x(103-97)
6A=1x3x5+1x3+3x5x7-3x5+5x7x9-3x5x7+7x9x11-5x7x9+...+99x101x103-99x101x97
6A=3+99x101x103
=>A=\(\frac{\text{3+99x101x103}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
\(B\text{=}\dfrac{3}{1\times3}+\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+...+\dfrac{3}{99\times101}\)
\(B\text{=}3\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(\dfrac{3-1}{1\times3}+\dfrac{5-3}{3\times5}+...+\dfrac{101-99}{99\times101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(B\text{=}\dfrac{3}{2}\times\left(1-\dfrac{1}{101}\right)\)
\(B\text{=}\dfrac{300}{202}\)