M= 1+(-2)+3+(-4)+...+2001+(-2002)+2003

   = 1+(-2+3)+(-4+5)+...+(-2002+2003)

   = 1 +   1   +   1    +...+      1            = 1002

       1002 số hạng

M=1+(-2)+3+(-4)+..........+2001+(-2002)+2003

M=[1+(-2)]+[3+(-4)+...........+[2001+(-2002)]+2003

M=(-1)+(-1)+...........+(-1)+2003{Có 1001 số (-1)}

M=(-1).1001+2003

M=(-1001)+2003

M=1002

=(1-2)+(3-4)+...+(2001-2002)+2003

=-1-1-1-...-1+2003  (1002 so -1)

=-1002+2003=1001

= 1001

KB ko

Đúng 100 % đấy

Từ 1 đến 2002 sẽ có:

\(\left(2002-1\right):1+1=2002\left(số\right)\)

=>Sẽ có 2002/2=1001 cặp có tổng là -1 là (1;-2);(3;-4);...;(2001;-2002)

M=1+(-2)+3+(-4)+...+2001+(-2002)+2003

=(1-2)+(3-4)+...+(2001-2002)+2003

=2003-1*1001

=2003-1001

=1002

Ta có : 

\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)

\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(A=3.\frac{6}{25}\)

\(A=\frac{18}{25}\)

Vậy \(A=\frac{18}{25}\)

Chúc bạn học tốt ~ 

\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)

\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)

\(=\frac{3.4.6}{25.4}\)

\(\Rightarrow A=\frac{18}{25}\)

A) =(2001+2002+2003)^3

=6006^3

B) =(2004+2005+2006)^3

=6015^3

C) =(2007+2008+2009)

=6024^3

2001³ + 2002³ + 2003³ + 2004³ + 2005³ + 2006³ + 2007³ + 2008³ + 2009³ =  \(\sum\limits^{2009}_{2001}x^3\) = 72541712030

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

nên x + 1 = 0 => x = -1

Vậy x = -1

b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)

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

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

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

Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)

nên 2004 + x = 0 => x = -2004

Vậy x = -2004

=))

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right).....\left(1-\dfrac{1}{2008^2}\right)\)

\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{2008^2-1}{2008^2}\)

\(=\dfrac{1.3}{4}.\dfrac{2.4}{9}.\dfrac{3.5}{16}....\dfrac{2007.2009}{2008^2}\)

\(=\left(\dfrac{1.2.3...2007}{2.3.4....2008}\right).\dfrac{3.4.5...2009}{2.3.4...2008}\)

\(=\dfrac{1}{2008}.\dfrac{2009}{2}=\dfrac{2009}{4016}\)