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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\text{A = 1.2.3 + 3.4.5+...99.100.101}\)
\(\text{A=1.3(5-3)+3.5(7-3)+}...+99.101\left(103-3\right)\)
\(=\left(1.3.5+3.5.7+5.7.9+...99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)
\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+...+99.101\right)\)
\(=13517400-3.171650\)
\(=13002450\)
D=1.2.3+3.4.5+...+99.100.101
D=1.2.3.4+5.6.7.4+........+99.100.101.4
D=1.2.3.4+5.6.7.(8-4)+........+99.100.101.(102-98)
D=(1.2.3.4+5.6.7.8+.........+99.100.101.102)-(1.2.3.4+5.6.7.8+....+98.99.100.101)
D=98.99.100.101
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 3/1.2.3 +3/2.3.4 + ............ + 3/98 . 99 . 100
2A = 2.3 / 1.2.3 + ...........+ 2.3/98.99.100
2A= 3. ( 2/1.2.3 + ............. + 2/98.99.100)
2A= 3.( 1/1.2 - 1/2.3 + .......... + 1/98 .99 - 1/99 . 100)
2A = 3.(1/2 - 1/990)
2A = 3. 247/495
2A = 741/495
A = 741/495 : 2
A = 247 / 330
bn cho mình gửi sắp đến thi học kì 2 rồi. đây là những món quà mà bn sẽ nhận đc:
1: áo quần
2: tiền
3: đc nhiều người yêu quý
4: may mắn cả
5: luôn vui vẻ trong cuộc sống
6: đc crush thích thầm
7: học giỏi
8: trở nên xinh đẹp
phật sẽ ban cho bn những điều này nếu cậu gửi tin nhắn này cho 25 người, sau 3 ngày bn sẽ có những đc điều đó. nếu bn ko gửi tin nhắn này cho 25 người thì bn sẽ luôn gặp xui xẻo, học kì 2 bn sẽ là học sinh yếu và bạn bè xa lánh( lời nguyền sẽ bắt đầu từ khi đọc) ( mình
cũng bị ép);-;
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=1.2.3+2.3.4+...+48.49.50\)
\(\Rightarrow4C=1.2.3.4+2.3.4.4+...+48.49.50.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+...+48.49.50.51-47.48.49.50\)
\(=48.49.50.51\)
\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
Nếu đúng thì k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{24}{1.2.3}+\frac{24}{2.3.4}+....+\frac{24}{19.20.21}\)
\(A=24.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{19.20.21}\right)\)
\(A=12.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{19.20.21}\right)\)
\(A=12.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-....-\frac{1}{20.21}\right)\)
\(A=12.\left(\frac{1}{2}-\frac{1}{420}\right)=12.\frac{209}{420}=\frac{209}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1.2.3+3.4.5+5.6.7+...+99.100.+101\)
\(A=1.3\left(5-3\right)+3.5\left(7-3\right)+5.7\left(9-3\right)+...+99.100\left(103-3\right)\)
\(=\left(1.3.5+3.5.7+5.7.9+99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)
\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+99.101\right)\)
\(=13517400-3.171650\)
\(=13002450\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=1.2.3+2.3.4+3.4.5+......+999.1000.1001
=>4A=1.2.3.4+2.3.4.4+3.4.5.4+.....+999.1000.1001.4
=>4A=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+......+999.1000.1001.(1002-998)
=>4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+.....+999.1000.1001.1002-998.999.1000.1001
=>4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....+999.1000.1001.1002-998.999.1000.1001
=>4A=999.1000.1001.1002
=>A=\(\frac{999.1000.1001.1002}{4}=.....\) (số này to quá,có thể bn ghi nhầm đề)
bạn ơi số to thì ko cần ghi kết quả cụ thể cx đc mà ko pải sai đề đâu bạn ấy lm đúng òi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{2015.2016.2017}\)
\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\frac{3}{2}.\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\frac{3}{2}.\left(\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\)
\(A=\frac{3}{4}-\frac{3}{2.2016.2017}< 1\)