1+(-2)+3+(-4)+...+2001+(-2002)

=[1+(-2)]+[3+(-4)]+...+[2001+(-2002)]

=(-1)+(-1)+...+(-1)                   (có 1001 số hạng)

=(-1).1001

=-1001

cám ơn

a/ S₁ = 1 + (-2) + 3 + (-4) + .. . + 2001 + ( -2002)

S₁ = [1 + (-2)] + [3 + (-4)] + .. . + [2001 + ( -2002)]

S₁ = (-1) + (-1) + ... + (-1)

2002 : 2 = 1001

S₁ = (-1) . 1001

S₁ = (-1001)

b/ S₂ = 1 + (-3) + 5 + (-7) + .. . + (-1999) + 2001

S₂ = [1 + (-3)] + [5 + (-7)] + .. . + [1997 + (-1999)] + 2001

S₂ = (-2) + (-2) + ... + (-2) + 2001

(1991 - 1) : 2 + 1 = 996 : 2 = 498

S₂ = (-2) . 498 + 2001

S₂ = (-996) + 2001

S₂ = 1005

c/ S₃ = 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + .. . + 1997 + (-1998) + (-1999) + 2000

S₃ = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 1997 + 1998 - 1999 - 2000

S₃ =(1 + 2 - 3 - 4)+(5 + 6 - 7 - 8)+ ... +(1997 + 1998 - 1999 - 2000)

S₃ = (-4) + (-4) + ... + (-4)

2000 : 4 = 500

S₃ = (-4) . 500

S₃ = -2000

hãy tính tổng S , biết : S = 1+2+ 3+....2000+2001+2002

a: S=(1-2)+(3-4)+...+(2001-2002)

=(-1)+(-1)+...+(-1)

=-1001

c: S=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+(1997-1998)

=-1

A=(1+2-3)+(-4+5+6-7)+(-8+9+10-11)+......(-2000+2001+2002-2003)

A=0+0....+0

A=0

Ta thấy 2-3-4=-5

            6-7-8=-9

           .............

           1998-1999-2000=-2001

=> 1+2-3-4+5+6-7-8+....-1999-2000+2001-2003=1-5+5-9+9-...-2001+2001+2002-2003

=> A= 1+2002-2003=0

Vậy A=0

\(=\left(1+2-3\right)+\left(-4+5+6-7\right)+...+\left(-2000+2001+2002-2003\right)\)

\(=0+0+0+...+0\)

\(=0\)

học tốt

S1 = 1 + (-2) + 3 + (-4) + ... + 2001 + (-2002)

= 1 - 2 + 3 - 4 + ... + 2001 - 2002

= (1 - 2) + (3 - 4) + ... + (2001 - 2002) (Có tất cả số cặp là: [(2002 - 1) : 1 + 1] : 2 = 1001 (cặp))

= (-1) + (-1) +...+ (-1) } 1001 chữ số (-1)

= (-1) . 1001 

= (-1001)

S2 = 1 + (-3) + 5 + (-7) +...+ (-1999) + 2001

= 1 - 3 + 5 - 7 + ... - 1999 + 2001

= (1 - 3) + (5 - 7) + ... (1997 - 1999) + 2001 (Có số cặp là: [(1999 - 1):2 + 1] : 2 =  500 (cặp))

= (-2) + (-2) + ... + (-2) + 2001 } 500 số (-2)

= (-2) . 500 + 2001

= -1000 + 2001

= 1001

= 

\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)

\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

\(=>x+2001=0\)

\(x=-2001\)

\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)

\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(=>x+1998=0\)

\(x=-1998\)

dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)

\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)

\(\Rightarrow x+2001=0\)

\(\Leftrightarrow x=-2001\)

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

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

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

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