Giải bài 6 trang 15 vở thực hành Toán 7

Đề bài

Tính:

a, \({\left( {1 + \frac{1}{2} - \frac{1}{4}} \right)^2}.\left( {2 + \frac{3}{7}} \right);\)

b, \(4:{\left( {\frac{1}{2} - \frac{1}{3}} \right)^3}.\)

Lời giải chi tiết

a,

          \(\begin{array}{l}{\left( {1 + \frac{1}{2} - \frac{1}{4}} \right)^2}.\left( {2 + \frac{3}{7}} \right)\\ = {\left( {\frac{4}{4} + \frac{2}{4} - \frac{1}{4}} \right)^2}.\left( {\frac{{14}}{7} + \frac{3}{7}} \right)\\ = {\left( {\frac{5}{4}} \right)^2}.\left( {\frac{{17}}{7}} \right)\\ = \frac{{25}}{{16}}.\frac{{17}}{7} = \frac{{425}}{{112}}.\end{array}\)

b,    

       \(\begin{array}{l}4:{\left( {\frac{1}{2} - \frac{1}{3}} \right)^3}\\ = 4:{\left( {\frac{3}{6} - \frac{2}{6}} \right)^3}\\ = 4:{\left( {\frac{1}{6}} \right)^3}\\ = {4.6^3}\\ = 4.216 = 864.\end{array}\)