Quy tắc nhân phân số: \(\frac{a}{b}. \frac{c}{d} = \frac{{a. c}}{{b. d}}\;\;(b \ne 0, d \ne 0);\;\;\;m. \frac{a}{b} = \frac{a}{b}. m = \frac{{a. m}}{b}\) Lưu ý. Giải chi tiết Giải bài 40 trang 40 sách bài tập Toán 6 – Cánh Diều Tập 2 – Bài 4. Phép nhân – phép chia phân số. Tính một cách hợp lí…
Đề bài/câu hỏi:
Tính một cách hợp lí
a) \(\frac{{11}}{4}.\frac{{ – 5}}{9}.\frac{8}{{33}};\)
b) \(\frac{{ – 5}}{6}.\frac{4}{{19}} + \frac{{ – 7}}{{12}}.\frac{4}{{19}} – \frac{{40}}{{57}}\)
c) \(\left( {\frac{{23}}{{41}} – \frac{{15}}{{82}}} \right).\frac{{41}}{{15}}\)
d) \(9\left( {\frac{{151515}}{{171717}} – \frac{{131313}}{{181818}}} \right);\)
e) \(\frac{{ – 13}}{8}.\left( {\frac{8}{{13}} + \frac{{32}}{{28}}} \right) – \frac{{15}}{7}\)
g) \(\frac{{{2^2}}}{{1.3}}.\frac{{{3^2}}}{{2.4}}.\frac{{{4^2}}}{{3.5}}.\frac{{{5^2}}}{{4.6}}.\frac{{{6^2}}}{{5.7}}\)
Hướng dẫn:
Quy tắc nhân phân số:
\(\frac{a}{b}.\frac{c}{d} = \frac{{a.c}}{{b.d}}\;\;(b \ne 0,d \ne 0);\;\;\;m.\frac{a}{b} = \frac{a}{b}.m = \frac{{a.m}}{b}\)
Lưu ý: Kết hợp với rút gọn phân số, tính chất phân phối, kết hợp, giao hoán.
Lời giải:
a) \(\frac{{11}}{4}.\frac{{ – 5}}{9}.\frac{8}{{33}} = \frac{{11}}{4}.\frac{8}{{33}}.\frac{{ – 5}}{9} = \frac{{11.8}}{{4.33}}.\frac{{ – 5}}{9} = \frac{{11.4.2}}{{4.11.3}}.\frac{{ – 5}}{9} = \frac{2}{3}.\frac{{ – 5}}{9} = \frac{{2.( – 5)}}{{3.9}} = \frac{{ – 10}}{{27}}\)
b)
\(\begin{array}{l}\frac{{ – 5}}{6}.\frac{4}{{19}} + \frac{{ – 7}}{{12}}.\frac{4}{{19}} – \frac{{40}}{{57}} = \left( {\frac{{ – 5}}{6} + \frac{{ – 7}}{{12}}} \right).\frac{4}{{19}} – \frac{{40}}{{57}} = \left( {\frac{{ – 10}}{{12}} + \frac{{ – 7}}{{12}}} \right).\frac{4}{{19}} – \frac{{40}}{{57}}\\ = \frac{{ – 17}}{{12}}.\frac{4}{{19}} – \frac{{40}}{{57}} = \frac{{ – 17.4}}{{12.9}} – \frac{{40}}{{57}} = \frac{{ – 17}}{{57}} – \frac{{40}}{{57}} = – 1\end{array}\)
c) \(\left( {\frac{{23}}{{41}} – \frac{{15}}{{82}}} \right).\frac{{41}}{{15}} = \frac{{23}}{{41}}.\frac{{41}}{{15}} – \frac{{15}}{{82}}.\frac{{41}}{{15}} = \frac{{23.41}}{{41.15}} – \frac{{15.41}}{{82.15}} = \frac{{23}}{{15}} – \frac{1}{2} = \frac{{46}}{{30}} – \frac{{15}}{{30}} = \frac{{31}}{{30}}\)
d)
\(\begin{array}{l}9.\left( {\frac{{151515}}{{171717}} – \frac{{131313}}{{181818}}} \right) = 9.\left( {\frac{{15.10101}}{{17.10101}} – \frac{{13.10101}}{{18.10101}}} \right) = 9.\left( {\frac{{15}}{{17}} – \frac{{13}}{{18}}} \right)\\ = 9.\frac{{15}}{{17}} – 9.\frac{{13}}{{18}} = \frac{{9.15}}{{17}} – \frac{{9.13}}{{9.2}} = \frac{{135}}{{17}} – \frac{{13}}{2} = \frac{{135.2 – 13.17}}{{17.2}} = \frac{{49}}{{34}}\end{array}\)
e)
\(\begin{array}{l}\frac{{ – 13}}{8}.\left( {\frac{8}{{13}} + \frac{{32}}{{28}}} \right) – \frac{{15}}{7} = \frac{{ – 13}}{8}.\frac{8}{{13}} + \frac{{ – 13}}{8}.\frac{{32}}{{28}} – \frac{{15}}{7}\\ = \frac{{ – 13.8}}{{8.13}} + \frac{{\left( { – 13} \right).32}}{{8.28}} – \frac{{15}}{7} = – 1 + \frac{{ – 13}}{7} – \frac{{15}}{7} = – 1 – \frac{{28}}{7} = – 5\end{array}\)
g) \(\frac{{{2^2}}}{{1.3}}.\frac{{{3^2}}}{{2.4}}.\frac{{{4^2}}}{{3.5}}.\frac{{{5^2}}}{{4.6}}.\frac{{{6^2}}}{{5.7}} = \frac{{{2^2}{{.3}^2}{{.4}^2}{{.5}^2}{{.6}^2}.}}{{1.3.2.4.3.5.4.6.5.7}} = \frac{{{2^2}{{.3}^2}{{.4}^2}{{.5}^2}{{.6}^2}.}}{{1.2.3.3.4.4.5.5.6.7}} = \frac{{2.6}}{7} = \frac{{12}}{7}\)