Giải bài 4 trang 30 SGK Toán 7 tập 1 - Cánh diều

Đề bài

Tính:

a)$0,3 - \frac{4}{9}:\frac{4}{3} \cdot \frac{6}{5} + 1$;

b)${\left( {\frac{{ - 1}}{3}} \right)^2} - \frac{3}{8}:{(0,5)^3} - \frac{5}{2} \cdot ( - 4)$;

c)$1 + 2:\left( {\frac{2}{3} - \frac{1}{6}} \right) \cdot ( - 2,25)$

d)$\left[ {\left( {\frac{1}{4} - 0,5} \right) \cdot 2 + \frac{8}{3}} \right]:2$.

Lời giải chi tiết

a)

$\begin{array}{l}0,3 - \frac{4}{9}:\frac{4}{3} \cdot \frac{6}{5} + 1\\ = \frac{3}{{10}} - \frac{4}{9}.\frac{3}{4}.\frac{6}{5} + 1\\ = \frac{3}{{10}} - \frac{2}{5} + 1\\ = \frac{3}{{10}} - \frac{4}{{10}} + \frac{{10}}{{10}}\\ = \frac{9}{{10}}\end{array}$

b)

$\begin{array}{l}{\left( {\frac{{ - 1}}{3}} \right)^2} - \frac{3}{8}:{(0,5)^3} - \frac{5}{2} \cdot ( - 4)\\ =\frac{1}{9} - \frac{3}{8}:(\frac{1}{2})^3 - \frac{5}{2}.\left( { - 4} \right)\\ = \frac{1}{9} - \frac{3}{8}:\frac{1}{8} - \frac{5}{2}.\left( { - 4} \right)\\ =\frac{1}{9} - \frac{3}{8}.8 - \frac{5}{2}.\left( { - 4} \right)\\ = \frac{1}{9} - 3 + 10\\ = \frac{1}{9} - \frac{{27}}{9} + \frac{{90}}{9}\\ = \frac{{64}}{9}\end{array}$

c)

$\begin{array}{l}1 + 2:\left( {\frac{2}{3} - \frac{1}{6}} \right) \cdot ( - 2,25)\\ = 1 + 2:\left( {\frac{4}{6} - \frac{1}{6}} \right) \cdot \left( { - \frac{225}{100}} \right)\\ = 1 + 2:\frac{1}{2}.\left( { - \frac{9}{4}} \right)\\ = 1 + 2.\frac{2}{1}.\left( { - \frac{9}{4}} \right)\\ = 1 + \left( { - 9} \right) =  - 8\end{array}$

d)

$\begin{array}{l}\left[ {\left( {\frac{1}{4} - 0,5} \right) \cdot 2 + \frac{8}{3}} \right]:2\\ = \left[ {\left( {\frac{1}{4} - \frac{2}{4}} \right) \cdot 2 + \frac{8}{3}} \right].\frac{1}{2}\\ = \left( {\frac{{ - 1}}{4}.2 + \frac{8}{3}} \right).\frac{1}{2}\\ = \left( {\frac{{ - 1}}{2} + \frac{8}{3}} \right).\frac{1}{2}\\= \left( {\frac{{ - 3}}{6} + \frac{16}{6}} \right).\frac{1}{2}\\ = \frac{{13}}{6}.\frac{1}{2} = \frac{{13}}{{12}}\end{array}$.