Giải bài 2 trang 25 SGK Toán 7 tập 1 - Cánh diều
Tính:
Đề bài
Tính
a) \(\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\)
b) \(1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\)
c)\(\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\)
d) \(0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)
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Lời giải chi tiết
a)
\(\begin{array}{l}\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\\ = \frac{{ - 1}}{5}.\frac{5}{3} - \frac{2}{3}.\frac{1}{2}\\ = \frac{{ - 1}}{3} - \frac{1}{3}\\ = \frac{{ - 2}}{3}\end{array}\)
b)
\(\begin{array}{l}1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\\ = 1 - {\left( {\frac{5}{9} - \frac{6}{9}} \right)^2}:\frac{4}{{27}}\\ = 1 - {\left( {\frac{{ - 1}}{9}} \right)^2}.\frac{{27}}{4}\\ = 1 - \frac{1}{{81}}.\frac{{27}}{4}\\ = 1 - \frac{1}{{12}}\\ = \frac{{11}}{{12}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\\ = \left[ {\left( {\frac{9}{{24}} - \frac{{10}}{{24}}} \right).6 + \frac{1}{3}} \right].4\\ = \left[ {\frac{{ - 1}}{{24}}.6 + \frac{1}{3}} \right].4\\ = \left[ {\frac{{ - 1}}{4} + \frac{1}{3}} \right].4\\ = \left[ {\frac{{ - 3}}{{12}} + \frac{4}{{12}}} \right].4\\ = \frac{1}{{12}}.4 = \frac{1}{3}\end{array}\)
d)
\(\begin{array}{l}0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\\ = \frac{4}{5}:\left\{ {\frac{1}{5} - 7.\left[ {\frac{1}{6} + \left( {\frac{{10}}{{42}} - \frac{{15}}{{42}}} \right)} \right]} \right\}\\ = \frac{4}{5}:\left\{ {\frac{1}{5} - 7.\left[ {\frac{7}{{42}} + \frac{{ - 5}}{{42}}} \right]} \right\}\\ = \frac{4}{5}:\left( {\frac{1}{5} - 7.\frac{2}{{42}}} \right)\\ = \frac{4}{5}:\left( {\frac{1}{5} - \frac{1}{3}} \right)\\ =\frac{4}{5}:\left( {\frac{3}{15} - \frac{5}{{15}}} \right)\\ = \frac{4}{5}:\frac{{ - 2}}{{15}}\\ = \frac{4}{5}.\frac{{ - 15}}{2}\\ = - 6\end{array}\)