Giải bài 6 trang 18 SBT Toán 7 Chân trời sáng tạo tập 1 - CTST

Phương pháp giải

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Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{{12}}{{23}}.\dfrac{7}{{13}} + \dfrac{{11}}{{23}}.\dfrac{7}{{13}} = \dfrac{7}{{13}}.\left( {\dfrac{{12}}{{23}} + \dfrac{{11}}{{23}}} \right) =\dfrac{7}{{13}}.\dfrac{23}{23}= \dfrac{7}{{13}}.1 = \dfrac{7}{{13}}\\b)\dfrac{4}{9}.\dfrac{{23}}{{11}} - \dfrac{1}{{11}}.\dfrac{4}{9} + \dfrac{4}{9} = \dfrac{4}{9}.\left( {\dfrac{{23}}{{11}} - \dfrac{1}{{11}} + 1} \right) = \dfrac{4}{9}.(2 + 1) = \dfrac{4}{9}.3 = \dfrac{4}{3}\end{array}\)

\(c)\left[ {\left( {\dfrac{{ - 5}}{7}} \right) + \dfrac{3}{5}} \right]:\dfrac{{2020}}{{2021}} + \left( {\dfrac{2}{5} - \dfrac{2}{7}} \right):\dfrac{{2020}}{{2021}}\)

\(\begin{array}{l} = \left[ {\left( {\dfrac{{ - 5}}{7}} \right) + \dfrac{3}{5}} \right].\dfrac{{2021}}{{2020}} + \left( {\dfrac{2}{5} - \dfrac{2}{7}} \right).\dfrac{{2021}}{{2020}}\\ = \left[ {\left( {\dfrac{{ - 5}}{7}} \right) + \dfrac{3}{5} + \dfrac{2}{5} - \dfrac{2}{7}} \right].\dfrac{{2021}}{{2020}}\end{array}\)

\(\begin{array}{l} = \left[ {\left( {\dfrac{{ - 5}}{7} - \dfrac{2}{7}} \right) + \left( {\dfrac{3}{5} + \dfrac{2}{5}} \right)} \right].\dfrac{{2021}}{{2020}}\\ = (-1 + 1).\dfrac{{2021}}{{2020}} = 0.\dfrac{{2021}}{{2020}} = 0\end{array}\)

\(\begin{array}{l}d)\dfrac{3}{8}:\left( {\dfrac{7}{{22}} - \dfrac{2}{{11}}} \right) + \dfrac{3}{8}:\left( {\dfrac{2}{5} - \dfrac{1}{{10}}} \right)\\ = \dfrac{3}{8}:\left( {\dfrac{7}{{22}} - \dfrac{4}{{22}}} \right) + \dfrac{3}{8}:\left( {\dfrac{4}{{10}} - \dfrac{1}{{10}}} \right)\\ = \dfrac{3}{8}:\dfrac{3}{{22}} + \dfrac{3}{8}:\dfrac{3}{{10}}\\ = \dfrac{3}{8}.\dfrac{{22}}{3} + \dfrac{3}{8}.\dfrac{{10}}{3}\\ = \dfrac{{22}}{8} + \dfrac{{10}}{8} = \dfrac{{32}}{8} = 4\end{array}\)

-- Mod Toán 7 HỌC247