Bài 6 trang 15 SBT Toán lớp 7 tập 1 Chân trời sáng tạo: Tính...

Tính

a) ${\left( {\dfrac{2}{5} - \dfrac{1}{3}} \right)^2}$

b) ${\left( {1\dfrac{1}{2} - 1,25} \right)^3}$

c) ${\left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)^2}:{\left( {1\dfrac{1}{2}} \right)^2}$

d) $2:{\left( {\dfrac{1}{2} - \dfrac{2}{3}} \right)^3}$

Tính biểu thức trong ngoặc trước —> Lũy thừa —> Nhân, chia

$\begin{array}{l}a){\left( {\dfrac{2}{5} - \dfrac{1}{3}} \right)^2} = {\left( {\dfrac{6}{{15}} - \dfrac{5}{{15}}} \right)^2}\\ = {\left( {\dfrac{1}{{15}}} \right)^2} = \dfrac{{1.1}}{{15.15}} = \dfrac{1}{{225}}\\b){\left( {1\dfrac{1}{2} - 1,25} \right)^3} = {\left( {\dfrac{3}{2} - \dfrac{5}{4}} \right)^3}\\ = {\left( {\dfrac{1}{4}} \right)^3} = \dfrac{1}{{64}}\\c){\left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)^2}:{\left( {1\dfrac{1}{2}} \right)^2}={\left( {\dfrac{3}{6} + \dfrac{2}{6}} \right)^2}:{\left( {\dfrac{3}{2}} \right)^2}\\ = {\left( {\dfrac{5}{6}} \right)^2}:{\left( {\dfrac{3}{2}} \right)^2}  = \dfrac{{25}}{{36}}:\dfrac{9}{4}= \dfrac{{25}}{{36}}.\dfrac{4}{9} = \dfrac{{25}}{{81}}\\d)2:{\left( {\dfrac{1}{2} - \dfrac{2}{3}} \right)^3} =2:{\left( {\dfrac{3}{6} - \dfrac{4}{6}} \right)^3}\\= 2:{\left( {\dfrac{{ - 1}}{6}} \right)^3}  = 2.[-{6^3}] \\= 2.( - 216) =  - 432\end{array}$