$Bài \ 1 : Tính \ \\ a) \ (0,125)^3 . 8^3\\= (0,125 . 8 )^3\\= 1^3\\= 1\\b) \ 2-\Bigg(\dfrac{-3}{2}\Bigg)+\dfrac{16}{4} : \dfrac{1}{2}\\ = \dfrac{2}{1}-\Bigg(\dfrac{-3}{2}\Bigg)+\dfrac{16}{4} . \dfrac{2}{1}\\=\dfrac{2}{1}.\Bigg[-\Bigg(\dfrac{-3}{2}\Bigg)+\dfrac{16}{4}\Bigg]\\=\dfrac{2}{1}.\Bigg(\dfrac{6}{4}+\dfrac{16}{4}\Bigg)\\=\dfrac{2}{1}.6\\=12\\ c) \ \dfrac{3^5.9}{3^7}\\=\dfrac{3^2.3^3.3^2}{3^3.3^2.3^2}\\=0\\d)\dfrac{3}{2}-\dfrac{5}{6}:\Bigg(\dfrac{1}{2}\Bigg)^2+\sqrt{4}\\=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{4}{1}+\dfrac{2}{1}\\=\dfrac{3}{2}-\dfrac{10}{3}+\dfrac{2}{1}\\=\dfrac{9}{6}-\dfrac{20}{6}+\dfrac{2}{1}\\=\dfrac{-11}{6}+\dfrac{2}{1}=\dfrac{-11}{6}+\dfrac{12}{6}\\=\dfrac{1}{6}$
