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\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)
Bài này dễ mà bn
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3.\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
=\(\left[6.\dfrac{1}{9}-\left(-1\right)+1\right]:\left(-\dfrac{4}{3}\right)\)
=\(\left[\dfrac{2}{3}-\left(-1\right)+1\right]:\left(-\dfrac{4}{3}\right)\)
=\(\dfrac{8}{3}:\left(-\dfrac{4}{3}\right)\)
=-2
a, \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{1\frac{25}{108}.320\frac{1}{25}+46\frac{3}{4}}{4\frac{16}{21}:\left(-1\frac{20}{21}\right)}=\frac{330\frac{1}{25}}{-2\frac{18}{41}}=\)\(-135,3164\)
a) \(\left(2-\frac{3}{2}\right)\left(2-\frac{4}{3}\right)\left(2-\frac{5}{4}\right)\left(2-\frac{6}{4}\right)\)
\(=\frac{1}{3}\left(-\frac{4}{3}+2\right)\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\left(-\frac{6}{4}+2\right)\)
\(=\frac{1.2.3\left(2-\frac{3}{2}\right)}{2.3.4}\)
\(=\frac{1.3\left(2-\frac{3}{2}\right)}{3.4}\)
\(=\frac{1.\left(2-\frac{3}{2}\right)}{4}\)
\(=\frac{2-\frac{3}{4}}{4}\)
\(=\frac{1}{2.4}\)
\(=\frac{1}{8}\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
\(=\frac{8028024\left(\frac{2003}{2004}+\frac{2004}{2003}\right)}{8028025}\)
\(=\frac{8028024.\frac{8028025}{4014012}}{8028025}\)
\(=\frac{16056050}{8028025}\)
= 2