Giải bài 1 trang 25

Đề bài

Tính

a) $\frac{1}{9}-0,3.\frac{5}{9}+\frac{1}{3};$

b) ${\left(\frac{-2}{3}\right)}^2+\frac{1}{6}-{\left(-0,5\right)}^3.$

Lời giải chi tiết

a)

$\begin{array}{c}\frac{1}{9}-0,3.\frac{5}{9}+\frac{1}{3}\\ =\frac{1}{9}-\frac{3}{10}.\frac{5}{9}+\frac{1}{3}\\ =\frac{1}{9}-\frac{3}{2.5}.\frac{5}{3.3}+\frac{1}{3}\\ =\frac{1}{9}-\frac{1}{6}+\frac{1}{3}\\ =\frac{2}{18}-\frac{3}{18}+\frac{6}{18}\\ =\frac{5}{18}\end{array}$

b)

$\begin{array}{c}{\left(\frac{-2}{3}\right)}^2+\frac{1}{6}-{\left(-0,5\right)}^3\\ =\frac{4}{9}+\frac{1}{6}-{\left(\frac{-1}{2}\right)}^3\\ =\frac{4}{9}+\frac{1}{6}-\left(\frac{-1}{8}\right)\\ =\frac{4}{9}+\frac{1}{6}+\frac{1}{8}\\ =\frac{32}{72}+\frac{12}{72}+\frac{9}{72}\\ =\frac{53}{72}\end{array}$