Ta có: A = 1+(-2)+3+(-4)+....+2003+(-2004) = 2005

     => A = (-1)+(-1)+(-1)+....+(-1) = (-1) x 2004 = -2004

a) (-124) + 24 = -100
b) 37. 78 + 37. 22
=  37 . ( 78 + 22 )
=  37 . 100
= 3700

\(a) (-124) + 24 = -100 b) 37. 78 + 37. 22 = 37 . ( 78 + 22 ) = 37 . 100 = 3700\)

a) (- 124) + 24                                          b) 37 . 78 + 37 . 22

= - ( 124 - 24)                                              =37 . ( 78 + 22)

= - 100                                                        =37 . 100 

                                                                    3700

a: \(\left(2345+45\right)-2345\)

\(=2345+45-2345\)

\(=45+\left(2345-2345\right)\)

=45+0

=45

b: \(\left(-2010\right)-\left(119-2010\right)\)

\(=-2010-119+2010\)

\(=\left(-2010+2010\right)-119\)

=0-119

=-119

c: \(\left(18+29\right)+\left(158-18-29\right)\)

\(=18+29+158-18-29\)

\(=158+\left(18-18\right)+\left(29-29\right)\)

=158+0+0

=158

d: \(126+\left(-20\right)+2004+\left(-106\right)\)

\(=126-20+2004-106\)

\(=\left(126-20-106\right)+2004\)

=0+2004

=2004

e: \(\left(-199\right)+\left(-200\right)+\left(201\right)\)

\(=-199-200+201\)

\(=-199+\left(201-200\right)\)

=-199+1

=-198

f: \(99+\left(-100\right)+101\)

\(=\left(99+101\right)-100\)

=200-100

=100

g: \(217+\left[43+\left(-217\right)+\left(-23\right)\right]\)

\(=217+43-217-23\)

\(=\left(217-217\right)+\left(43-23\right)\)

=0+20

=20

\(=64\cdot\dfrac{-1}{8}+100\cdot\dfrac{1}{100}\)

=-8+1

=-7

\(A=-\dfrac{1}{6}-\left[\dfrac{9}{16}-\dfrac{9}{8}+\dfrac{3}{4}\right]\)

\(=\dfrac{-1}{6}-\dfrac{9-18+12}{16}=\dfrac{-1}{6}-\dfrac{3}{16}\)

\(=\dfrac{-8}{48}-\dfrac{9}{48}=-\dfrac{17}{48}\)

\(B=\left(-\dfrac{3}{4}+\dfrac{3}{4}\right)^2+\dfrac{3}{4}\cdot\dfrac{-5}{9}+7^3\)

=-15/36+343

=343-5/12=4111/12

-2008 - 2006 - 2004 - ..... - 2

= -(2 + 4 + 6 + ...... + 2008)

= - (2 + 2008) x 1004

= -2018040

\(\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}=\dfrac{5\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}{10\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}=\dfrac{1}{2}\)