Giải bài 4 trang 27 SGK Toán 7 tập 1 – Chân trời sáng tạo

Đề bài

Tính giá trị các biểu thức sau:

a)\(A = \left[ {\left( { – 0,5} \right) – \frac{3}{5}} \right]:\left( { – 3} \right) + \frac{1}{3} – \left( { – \frac{1}{6}} \right):\left( { – 2} \right)\)

b)\(B = \left( {\frac{2}{{25}} – 0,036} \right):\frac{{11}}{{50}} – \left[ {\left( {3\frac{1}{4} – 2\frac{4}{9}} \right)} \right].\frac{9}{{29}}\)

Lời giải chi tiết

a)

\(\begin{array}{l}A = \left[ {\left( { – 0,5} \right) – \frac{3}{5}} \right]:\left( { – 3} \right) + \frac{1}{3} – \left( { – \frac{1}{6}} \right):\left( { – 2} \right)\\ = \left( {\frac{{ – 5}}{{10}} – \frac{6}{{10}}} \right).\frac{{ – 1}}{3} + \frac{1}{3} – \frac{-1}{6}.\frac{{ – 1}}{2}\\ = \frac{{ – 11}}{{10}}.\frac{{ – 1}}{3} + \frac{1}{3} – \frac{1}{12}\\ = \frac{{11}}{{30}} + \frac{1}{3} + \frac{{ – 1}}{{12}}\\ = \frac{{22}}{{60}} + \frac{{20}}{{60}} – \frac{{ 5}}{{60}}\\ = \frac{{37}}{{60}}\end{array}\)

b)

\(\begin{array}{l}B = \left( {\frac{2}{{25}} – 0,036} \right):\frac{{11}}{{50}} – \left[ {\left( {3\frac{1}{4} – 2\frac{4}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{2}{{25}} – \frac{{36}}{{1000}}} \right).\frac{{50}}{{11}} – \left[ {\left( {\frac{{13}}{4} – \frac{{22}}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{{10}}{{125}} – \frac{9}{{250}}} \right).\frac{{50}}{{11}} – \left[ {\left( {\frac{{117}}{{36}} – \frac{{88}}{{36}}} \right)} \right].\frac{9}{{29}}\\ = \frac{{ 11}}{{250}}.\frac{{50}}{{11}} – \frac{{29}}{{36}}.\frac{9}{{29}}\\ = \frac{{ 1}}{{5}} – \frac{1}{4}\\ = \frac{{ 4}}{{20}} – \frac{{5}}{{20}}\\ = \frac{{ – 1}}{{20}}\end{array}\)