Tính: a) 1/9 - 0,3.5/9 + 1/3 b) (-2/3)^2 + 1/6 - (-0,5)^3

* Hướng dẫn giải

$$\begin{gathered} a) \frac{1}{9}-0,3.\frac{5}{9}+\frac{1}{3} \\ =\frac{1}{9}-\frac{3}{10}.\frac{5}{9}+\frac{1}{3} \\ =\frac{1}{9}-\frac{1}{6}+\frac{1}{3} \\ =\frac{2}{18}-\frac{3}{18}+\frac{6}{18}=\frac{5}{18} \\ b) {\left(\frac{{\displaystyle -2}}{{\displaystyle 3}}\right)}^2+\frac{{\displaystyle 1}}{{\displaystyle 6}}-{(-0,5)}^3 \\ =\frac{{(-2)}^2}{3^2}+\frac{1}{6}-{\left(\frac{-1}{2}\right)}^3 \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle {(-1)}^3}}{{\displaystyle 2^3}} \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle -1}}{{\displaystyle 8}} \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}+\frac{{\displaystyle 1}}{{\displaystyle 8}} \\ =\frac{{\displaystyle 32}}{{\displaystyle 72}}+\frac{{\displaystyle 12}}{{\displaystyle 72}}+\frac{{\displaystyle 9}}{{\displaystyle 72}}=\frac{{\displaystyle 53}}{{\displaystyle 72}} \end{gathered}$$