a) \(1-2-3+4+5-6-7+...+2001-2002-2003+2004\)

  \(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2001-2002-2003+2004\right)\)

  \(=0+0+...+0=0\)

b) \(1+2-3-4+5+6-7-8+...+2001+2002-2003-2004\)

   \(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2001+2002-2003-2004\right)\)

   \(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

   \(=\left(-4\right)\cdot501=\left(-2004\right)\)

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

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

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

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

ko bit

Ko biết

a) 1-2-3+4+5-6-7+8+...+2001-2002-2003+2004 

S = (1+2-3+4) + (5+6-7-8) + ... + (2001+2002-2003-2004) + (2005+2006)

S = (-4) + (-4) + ... + (-4) + (2005+2006)

dãy S có 2004 - 1 : 1 + 1 = 2004 số hạng

dãy S có 2004 : 4 = 501 chữ số (-4)

do đó S = -4. 501 = -2004

S = -2004 + (2005+2006)

S = -2004 + 4011

S = 2007

b) tương tự nhé!!

675676587689689

a) Nhóm 4 số hạng liên tiếp từ đầu dãy:

A = (1-2-3+4)+(5-6-7+8)+(9-10-11+12)+ ...+(2001-2002-2003+2004) = 0

b) Nhóm 4 số hạng liên tiếp bắt đầu từ số thứ 2:

B = 1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006 = 1+2006 = 2007.

chết cho mk xin lỗi mk làm câu b) mà kéo nhầm câu a đó bn!!

sorry nhìu!! 654647567689