Ta áp dụng các quy tắc cộng, trừ, nhân. Trả lời Giải Bài 4 trang 10 sách bài tập toán 7 tập 1 – Chân trời sáng tạo – Bài 2: Các phép tính với số hữu tỉ. Tính…
Đề bài/câu hỏi:
Tính
a) \(\dfrac{6}{7}.\left( { – \dfrac{1}{8}} \right) + \dfrac{6}{7}.\left( { – \dfrac{3}{4}} \right)\)
b) \(\left( {\dfrac{{ – 7}}{{17}}} \right).\dfrac{5}{{12}} + \left( {\dfrac{{ – 7}}{{17}}} \right).\dfrac{7}{{12}} + \left( {\dfrac{{ – 10}}{{17}}} \right)\)
c) \(\left[ {\dfrac{3}{5} + \left( {\dfrac{{ – 1}}{4}} \right)} \right]:\dfrac{3}{7} + \left[ {\left( {\dfrac{{ – 3}}{4}} \right) + \dfrac{2}{5}} \right]:\dfrac{3}{7}\)
d) \(\dfrac{7}{8}:\left( {\dfrac{2}{9} – \dfrac{1}{{18}}} \right) + \dfrac{7}{8}:\left( {\dfrac{1}{{36}} – \dfrac{5}{{12}}} \right)\)
Hướng dẫn:
Ta áp dụng các quy tắc cộng, trừ, nhân, chia các phân số
Lời giải:
\(a)\dfrac{6}{7}.\left( { – \dfrac{1}{8}} \right) + \dfrac{6}{7}.\left( { – \dfrac{3}{4}} \right) = \dfrac{6}{7}.\left( { – \dfrac{1}{8} – \dfrac{3}{4}} \right)\\ = \dfrac{6}{7}.\left( { – \dfrac{1}{8} – \dfrac{6}{8}} \right) = \dfrac{6}{7}.\left( { – \dfrac{7}{8}} \right) = – \dfrac{6}{8} = – \dfrac{3}{4}\\b)\left( {\dfrac{{ – 7}}{{17}}} \right).\dfrac{5}{{12}} + \left( {\dfrac{{ – 7}}{{17}}} \right).\dfrac{7}{{12}} + \left( {\dfrac{{ – 10}}{{17}}} \right)\\ = \left( {\dfrac{{ – 7}}{{17}}} \right).\left( {\dfrac{5}{{12}} + \dfrac{7}{{12}}} \right) + \left( {\dfrac{{ – 10}}{{17}}} \right)\\ = \left( {\dfrac{{ – 7}}{{17}}} \right).1 + \left( {\dfrac{{ – 10}}{{17}}} \right)\\ =\dfrac{-7}{17}+\dfrac{-10}{17}= \dfrac{{ – 17}}{{17}} = – 1\\c)\left[ {\dfrac{3}{5} + \left( {\dfrac{{ – 1}}{4}} \right)} \right]:\dfrac{3}{7} + \left[ {\left( {\dfrac{{ – 3}}{4}} \right) + \dfrac{2}{5}} \right]:\dfrac{3}{7}\\ = \left( {\dfrac{{12}}{{20}} + \dfrac{-5}{{20}}} \right):\dfrac{3}{7} + \left( {\dfrac{{ – 15}}{{20}} + \dfrac{8}{{20}}} \right):\dfrac{3}{7}\\ = \dfrac{7}{{20}}:\dfrac{3}{7} + \dfrac{{ – 7}}{{20}}:\dfrac{3}{7}\\ = \left( {\dfrac{7}{{20}} + \dfrac{{ – 7}}{{20}}} \right):\dfrac{3}{7} = 0.\dfrac{7}{3} = 0\\d)\dfrac{7}{8}:\left( {\dfrac{2}{9} – \dfrac{1}{{18}}} \right) + \dfrac{7}{8}:\left( {\dfrac{1}{{36}} – \dfrac{5}{{12}}} \right)\\ = \dfrac{7}{8}:\left( {\dfrac{4}{{18}} – \dfrac{1}{{18}}} \right) + \dfrac{7}{8}:\left( {\dfrac{1}{{36}} – \dfrac{{15}}{{36}}} \right)\\ = \dfrac{7}{8}: {\dfrac{3}{18}} + \dfrac{7}{8}:{\dfrac{{ – 14}}{{36}}}\\= \dfrac{7}{8}: {\dfrac{1}{6}} + \dfrac{7}{8}:{\dfrac{{ – 7}}{{18}}}\\ = \dfrac{7}{8}.6 + \dfrac{7}{8}.\dfrac{{ – 18}}{7} = \dfrac{{42}}{8} + \dfrac{{ – 18}}{8} = \dfrac{{24}}{8} = 3\)