Question

tính :

a, (2-\(\dfrac{3}{2}\) ) . (2-\(\dfrac{4}{3}\) ) . (2-\(\dfrac{5}{4}\) ) . (2-\(\dfrac{6}{5}\) )

b,\(\dfrac{1}{2002}\) + \(\dfrac{2003.2001}{2002}\) - 2003

c, (\(\dfrac{2003}{2004}\) + \(\dfrac{2004}{2003}\) ) : \(\dfrac{8028025}{8028024}\)

giúp mình với mình đang cần gấp

Answer 1

a, \(\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)

\(=\left(\dfrac{4}{2}-\dfrac{3}{2}\right)\left(\dfrac{6}{3}-\dfrac{4}{3}\right)\left(\dfrac{8}{4}-\dfrac{5}{4}\right)\left(\dfrac{10}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

Answer 2

b. \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)\(=\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-\dfrac{2003.2002}{2002}\) = \(\dfrac{1+2003.2001-2003.2002}{2002}\) = \(\dfrac{1+\left(2003\left(2001-2002\right)\right)}{2002}\) = \(\dfrac{1+2003.\left(-1\right)}{2002}\) = \(\dfrac{1+\left(-2003\right)}{2002}\) = \(\dfrac{-2002}{2002}=-1\)

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