Bài tập 18 trang 27 Tài liệu dạy – học Toán 7 tập 1

Đề bài

Tính

a) \({3 \over 5}.\left( {{5 \over 3} – {2 \over 7}} \right) – \left( {{7 \over 3} – {3 \over 7}} \right).{3 \over 5}\)

b) \({8 \over 9}:\left( {{3 \over 5} – {2 \over 7}} \right) – {8 \over 9}:\left( {{6 \over 7} – {4 \over 5}} \right)\)

Lời giải chi tiết

\(\eqalign{  & a){3 \over 5}.\left( {{5 \over 3} – {2 \over 7}} \right) – \left( {{7 \over 3} – {3 \over 7}} \right).{3 \over 5} = {3 \over 5}.\left( {{5 \over 3} – {2 \over 7} – {7 \over 3} + {3 \over 7}} \right)  \cr  &  = {3 \over 5}.\left[ {\left( {{5 \over 3} – {7 \over 3}} \right) + \left( { – {2 \over 7} + {3 \over 7}} \right)} \right]  \cr  &  = {3 \over 5}\left( {{{ – 2} \over 7} + {1 \over 7}} \right) = {3 \over 5}.\left( {{{ – 14} \over {12}} + {3 \over {21}}} \right) = {3 \over 5}.\left( {{{ – 11} \over {21}}} \right) = {{ – 11} \over {35}} \cr} \)

\(\begin{array}{l}b)\frac{8}{9}:\left( {\frac{3}{5} – \frac{2}{7}} \right) – \frac{8}{9}:\left( {\frac{6}{7} – \frac{4}{5}} \right)\\ = \frac{8}{9}:\left( {\frac{{21}}{{35}} – \frac{{10}}{{35}}} \right) – \frac{8}{9}:\left( {\frac{{30}}{{35}} – \frac{{28}}{{35}}} \right)\\ = \frac{8}{9}:\frac{{11}}{{35}} – \frac{8}{9}:\frac{2}{{35}}\\ = \frac{8}{9}.\frac{{35}}{{11}} – \frac{8}{9}.\frac{{35}}{2}\\ = \frac{8}{9}.\left( {\frac{{35}}{{11}} – \frac{{35}}{2}} \right)\\ = \frac{8}{9}.\left( {\frac{{70}}{{22}} – \frac{{385}}{{22}}} \right)\\ = \frac{8}{9}.\frac{{ – 315}}{{22}}\\ = \frac{{2.4}}{9}.\frac{{ – 35.9}}{{2.11}}\\ = \frac{{ – 140}}{{11}}\end{array}\)

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