Answer:
Step-by-step explanation:
[tex](-1)^{n} = -1 , if \ n \ is \ a \ odd \ integer\\\\\\\\(-1)^{2003} =-1 \\\\(\dfrac{-3}{4)}^{2}=\dfrac{3}{4}, \ as \ 2 \ is \ even \ number.\\\\(\dfrac{2}{3})^{3}*(\dfrac{-3}{4})^{2}*(-1)^{2003}=\dfrac{2^{3}}{3^{3}}*\dfrac{3^{2}}{4^{2}}*(-1)\\\\\\=\dfrac{2^{3}}{3^{3}}*\dfrac{3^{2}}{2^{4}}*(-1)\\\\\\=\dfrac{-1}{3^{3-2}*2^{4-3}}\\\\\\=\dfrac{-1}{3^{1}*2^{1}}\\\\=\dfrac{-1}{6}[/tex]
[tex](\dfrac{2}{5})^{2}*(\dfrac{-5}{12})^{3}=\dfrac{2^{2}}{5^{2}}*\dfrac{(-5)^{3}}{(2^{2}*3)^{3}}\\\\\\=\dfrac{2^{2}}{5^{2}}*\dfrac{- 5^{3}}{2^{6}*3^{3}}\\\\\\= -\dfrac{5^{3-2}}{2^{6-2}*3^{3}}\\\\\\= -\dfrac{5}{2^{4}3^{3}}\\\\= -\dfrac{5}{16*3}\\\\\\= \dfrac{-5}{48}[/tex][tex]HINT: \dfrac{a^{m}}{a^{n}}=a^{m-n}, m >n\\\\\\\dfrac{a^{m}}{a^{n}}=\dfrac{1}{a^{n-m}}, n >m\\\\\\(a^{m})^{n}=a^{m*n}[/tex]
