Giải bài 4 trang 10 SBT Toán 7 Chân trời sáng tạo tập 1 - CTST

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Lời giải chi tiết

\(a)\dfrac{6}{7}.\left( { - \dfrac{1}{8}} \right) + \dfrac{6}{7}.\left( { - \dfrac{3}{4}} \right) = \dfrac{6}{7}.\left( { - \dfrac{1}{8} - \dfrac{3}{4}} \right)\\ = \dfrac{6}{7}.\left( { - \dfrac{1}{8} - \dfrac{6}{8}} \right) = \dfrac{6}{7}.\left( { - \dfrac{7}{8}} \right) =  - \dfrac{6}{8} =  - \dfrac{3}{4}\\b)\left( {\dfrac{{ - 7}}{{17}}} \right).\dfrac{5}{{12}} + \left( {\dfrac{{ - 7}}{{17}}} \right).\dfrac{7}{{12}} + \left( {\dfrac{{ - 10}}{{17}}} \right)\\ = \left( {\dfrac{{ - 7}}{{17}}} \right).\left( {\dfrac{5}{{12}} + \dfrac{7}{{12}}} \right) + \left( {\dfrac{{ - 10}}{{17}}} \right)\\ = \left( {\dfrac{{ - 7}}{{17}}} \right).1 + \left( {\dfrac{{ - 10}}{{17}}} \right)\\ =\dfrac{-7}{17}+\dfrac{-10}{17}= \dfrac{{ - 17}}{{17}} =  - 1\\c)\left[ {\dfrac{3}{5} + \left( {\dfrac{{ - 1}}{4}} \right)} \right]:\dfrac{3}{7} + \left[ {\left( {\dfrac{{ - 3}}{4}} \right) + \dfrac{2}{5}} \right]:\dfrac{3}{7}\\ = \left( {\dfrac{{12}}{{20}} - \dfrac{5}{{20}}} \right):\dfrac{3}{7} + \left( {\dfrac{{ - 15}}{{20}} + \dfrac{8}{{20}}} \right):\dfrac{3}{7}\\ = \dfrac{7}{{20}}:\dfrac{3}{7} + \dfrac{{ - 7}}{{20}}:\dfrac{3}{7}\\ = \left( {\dfrac{7}{{20}} + \dfrac{{ - 7}}{{20}}} \right):\dfrac{3}{7} = 0.\dfrac{7}{3} = 0\\d)\dfrac{7}{8}:\left( {\dfrac{2}{9} - \dfrac{1}{{18}}} \right) + \dfrac{7}{8}:\left( {\dfrac{1}{{36}} - \dfrac{5}{{12}}} \right)\\ = \dfrac{7}{8}:\left( {\dfrac{4}{{18}} - \dfrac{1}{{18}}} \right) + \dfrac{7}{8}:\left( {\dfrac{1}{{36}} - \dfrac{{15}}{{36}}} \right)\\ = \dfrac{7}{8}: {\dfrac{3}{18}} + \dfrac{7}{8}:{\dfrac{{ - 14}}{{36}}}\\= \dfrac{7}{8}: {\dfrac{1}{6}} + \dfrac{7}{8}:{\dfrac{{ - 7}}{{18}}}\\ = \dfrac{7}{8}.6 + \dfrac{7}{8}.\dfrac{{ - 18}}{7} = \dfrac{{42}}{8} + \dfrac{{ - 18}}{8} = \dfrac{{24}}{8} = 3\) 

-- Mod Toán 7 HỌC247