\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)

    \(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)

     \(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)

      \(=\frac{1}{2003}\)

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

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

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

(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

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

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

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

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

1001/ 2004 nhé chúc bạn học giỏi

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)

\(A=\frac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)

\(A=\frac{1}{2004}\)

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)

\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)

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

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

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

\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)

~ Hok tốt ~

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

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

\(=\left(-\frac{2}{3}+2\right):\left(-\frac{4}{3}\right)\)

\(=\frac{4}{3}:\left(-\frac{4}{3}\right)=-1\)

dễ mà

trừ đi rùi rút gọn

\(C=\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right).\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{1}{12}\right)\)

\(C=\left(\frac{9}{15}-\frac{4}{15}\right).\left(\frac{4}{14}-\frac{3}{14}\right)-\left(\frac{15}{27}-\frac{7}{27}\right).\left(\frac{5}{5}-\frac{3}{5}\right)+\left(\frac{12}{12}-\frac{11}{12}\right).\left(\frac{12}{12}+\frac{1}{12}\right)\)

\(C=\frac{5}{15}.\frac{1}{14}.\frac{8}{27}.\frac{2}{5}.\frac{1}{12}.\frac{13}{12}\)

\(C=\frac{5.1.8.2.1.13}{15.14.27.5.12.12}\)

\(C=\frac{5.2.4.2.13}{3.5.14.27.5.4.3.2.2.3}\)

\(C=\frac{13}{3.14.27.5.3.3}\)

\(C=\frac{13}{51030}\)