\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{2002}\right).\left(1+\frac{1}{2003}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2003}{2002}.\frac{2004}{2003}\)

\(=\frac{2004}{2}=1002\)

(1+1/2)(1+1/3)(1+1/4)...(1+1/2003)=3/2.4/3.5/4.....2004/2003=3.4.5.....2004/2.3.4.....2003=2004/2=1002

\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2002}\right).\left(1+\frac{1}{2003}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2003}{2002}.\frac{2004}{2003}\)

\(=\frac{3.4.5...2003.2004}{2.3.4...2002.2003}\)

\(=\frac{2004}{2}=1002\)

\(\left[6.\left(-\dfrac{1}{3}\right)^2-3.\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)

=\(\left[6.\dfrac{1}{9}-\left(-1\right)+1\right]:\left(-\dfrac{4}{3}\right)\)

=\(\left[\dfrac{2}{3}-\left(-1\right)+1\right]:\left(-\dfrac{4}{3}\right)\)

=\(\dfrac{8}{3}:\left(-\dfrac{4}{3}\right)\)

=-2

Bài 1:

1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3

= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3

= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3 

= \(\frac{11}{10}\)- 3

= \(\frac{-19}{10}\)

2) \(\frac{5}{6}\):  \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)

= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)

= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))

= \(\frac{5}{6}\).( -30)

= -25

mách mình mấy câu kia với

Gộp nhóm 4 => A = -4 * 500+2001+2002-2003=0

B = 

a) tạm bỏ số 1 ra => có 2012 số hạng=> có 1006 cặp =(-1)

=> A=1+-(-1).1006=-1005