- Thực hiện các phép tính sau:
a)\(\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{4}\right)\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
K
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27 tháng 7 2017
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)
Bài này dễ mà bn
23 tháng 8 2018
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
TH
9 tháng 8 2017
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)
\(A=\frac{1}{2004}\)
3 tháng 6 2019
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
TN
12 tháng 5 2016
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
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