Thực hiện phép tính (hợp lý nếu có thể):
b)2004 + [ 520 + (-2004)]
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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: \(\left(2345+45\right)-2345\)
\(=2345+45-2345\)
\(=45+\left(2345-2345\right)\)
=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
\(\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}=\dfrac{5\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}{10\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}=\dfrac{1}{2}\)
\(b)\)\(2004+\left[520+\left(-2004\right)\right]\)
\(=\)\(2004+520-2004\)
\(=\)\(\left(2004-2004\right)+520\)
\(=\)\(0+520\)
\(=\)\(520\)