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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
![](https://rs.olm.vn/images/avt/0.png?1311)
ta co : 13 + 23 +...+ 103 = 3025
=>2.13 + 2.23+...+2.103 = 2.3025
=> 26 + 46 +...+ 206 =6050
=> [23 + 3] + [ 43 + 3] + ...+ [203 + 3] =6050
=> 23 + 43 +... 203 = 6050 - 3.10
=> S = 6020
Ta có :
13 + 25 + ... + 103 = 3025
=> 2,13 + 2,23 + .... + 2,103 = 2,3025
=> 26 + 46 + .... + 206 = 6050
=> ( 23 + 3 ) + ( 43 + 3 ) + ..... + ( 203 + 3 ) = 6050
=> 23 + 43 + .... 203 = 6050 - 3,10
=> S = 6020
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 13 + 23 + 33 +...+103 = 3025
⇒ 2.13 + 2.23 +..+ 2.103 = 2.3025
⇒ 26 + 46 +..+206 = 6050
⇒ (23+3) + (43+3) +..+(203+3) = 6050
⇒ 23 + 43 +..+203 = 6050 - 3.10
⇒ S = 6020
ta có: 13 + 23 + 33 +...+103 = 3025
⇒ 2.13 + 2.23 +..+ 2.103 = 2.3025
⇒ 26 + 46 +..+206 = 6050
⇒ (23+3) + (43+3) +..+(203+3) = 6050
⇒ 23 + 43 +..+203 = 6050 - 3.10
⇒ S = 6020
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
Đặt A= 13+23+...+103= 3025
2A=26+46+...+206
2A=(23+3)+(43+3)+...+(203+3)=6050
S=6050-(3.10)=6020
Ko hiểu cứ hỏi mình
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1^3+2^3+3^3+...+10^3=3025\\ S=2^3+4^3+6^3+...+20^3\\ =\left(2\cdot1\right)^3+\left(2\cdot2\right)^3+\left(2\cdot3\right)^3+...+\left(2\cdot10\right)^3\\ =2^3\cdot1^3+2^3\cdot2^3+2^3\cdot3^3+...+2^3\cdot10^3\\ =2^3\left(1^3+2^3+3^3+...+10^3\right)\\ =8\cdot3025\\ =24200\)
Ta có:
\(S=2^3+4^3+6^3+...+20^3.\)
\(\Rightarrow S=2^3.1^3+2^3.2^3+2^3.3^3+...+2^3.10^3.\)
\(\Rightarrow S=2^3\left(1^3+2^3+3^3+...+10^3\right).\)
\(\Rightarrow S=8.3025=24200.\)
Vậy \(S=24200.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 13+10+23+20+33+30+...+103+100
S = 13+23+33+...+103+10.100
S = 3025+1000
S = 4025
![](https://rs.olm.vn/images/avt/0.png?1311)
=51005000
chúc bn hk tốt!!