Giải bài 7 trang 21

Đề bài

Tính:

a)${\left(\frac{2}{5}+\frac{1}{2}\right)}^2$;                 b)${\left(0,75-1\frac{1}{2}\right)}^3;$

c)${\left(\frac{3}{5}\right)}^{15}:{\left(0,36\right)}^5$;         d)${\left(1-\frac{1}{3}\right)}^8:{\left(\frac{4}{9}\right)}^3$

Lời giải chi tiết

a)${\left(\frac{2}{5}+\frac{1}{2}\right)}^2={\left(\frac{4}{10}+\frac{5}{10}\right)}^2={\left(\frac{9}{10}\right)}^2=\frac{81}{100}$;               

 b)${\left(0,75-1\frac{1}{2}\right)}^3={\left(\frac{3}{4}-\frac{3}{2}\right)}^3={\left(\frac{3}{4}-\frac{6}{4}\right)}^3={\left(-\frac{3}{4}\right)}^3=\frac{-27}{64};$

c)

$\begin{array}{c}{\left(\frac{3}{5}\right)}^{15}:{\left(0,36\right)}^5={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{9}{25}\right)}^5\\ ={\left(\frac{3}{5}\right)}^{15}:{\left[{\left(\frac{3}{5}\right)}^2\right]}^5={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{3}{5}\right)}^{10}={\left(\frac{3}{5}\right)}^5\end{array}$

d) ${\left(1-\frac{1}{3}\right)}^8:{\left(\frac{4}{9}\right)}^3={\left(\frac{3}{3}-\frac{1}{3}\right)}^8:{\left((\frac{2}{3}{)}^2\right)}^3={\left(\frac{2}{3}\right)}^8:{\left(\frac{2}{3}\right)}^6={\left(\frac{2}{3}\right)}^{8-6}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$