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

1. \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2003}+\dfrac{2}{2004}-\dfrac{2}{2005}}{\dfrac{3}{2003}+\dfrac{3}{2004}-\dfrac{3}{2005}}\)

=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{5\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}-\)\(\dfrac{2\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}{3\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}\)

=\(\dfrac{1}{5}-\dfrac{2}{3}\)

=\(-\dfrac{7}{15}\)

9 bạn ạ

mình nhanh nhất

bai toan nay qua don gian

là 9 nhé , mình trước nhé 

\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times63}\)

\(A=\frac{7}{10}.\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{23}\right)+....+\left(\frac{1}{53}-\frac{1}{63}\right)\right]\)

\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{53}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}.\frac{20}{63}\)

\(A=\frac{2}{9}\)

Ta có:S=23+43+63+…+203

=>S=23+43+63+…+203+3+3+3+…+3-3-3-3-…-3

=>S=(23+3)+(43+3)+(63+3)+…+(203+3)-(3+3+3+…+3)

=>S=26+46+66+…+206-(3+3+3+…+3)

Từ 26 đến 206 có:

                  (206-26):20+1=10(số)

=>S=26+46+66+…+206-3.10

=>S=2.(13+23+33+…+103)-30

mà 13+23+33+…+103=3025

=>S=2.3025-30

=>S=6050-30

=>S=6020

Vậy S=6020.

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

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

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 :

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

đề sai nk bn

\(1^3+2^3+3^3+...+10^3=3025\\ S=2^3+4^3+6^3+...+20^3\\ =\left(2\cdot1\right)^3+\left(2\cdot2\right)^3+\left(2\cdot3\right)^3+...+\left(2\cdot10\right)^3\\ =2^3\cdot1^3+2^3\cdot2^3+2^3\cdot3^3+...+2^3\cdot10^3\\ =2^3\left(1^3+2^3+3^3+...+10^3\right)\\ =8\cdot3025\\ =24200\)

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 \(S=24200.\)