Bài 40 trang 40 SBT Toán 6 – Cánh Diều Tập 2: Tính một cách hợp lí \(\frac{{11}}{4}. \frac{{ - 5}}{9}. \frac{8}{{33}};\) \(\frac{{ - 5}}{6}. \frac{4}{{19}} + \frac{{ - 7}}{{12}}...

Đề bài :

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 giải :

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 chi tiết :

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) \(\frac{{{2^2}}}{{1.3}}.\frac{{{3^2}}}{{2.4}}.\frac{{{4^2}}}{{3.5}}.\frac{{{5^2}}}{{4.6}}.\frac{{{6^2}}}{{5.7}} = \frac{{{2^2}{{.3}^2}{{.4}^2}{{.5}^2}{{.6}^2}.}}{{1.3.2.4.3.5.4.6.5.7}} = \frac{{{2^2}{{.3}^2}{{.4}^2}{{.5}^2}{{.6}^2}.}}{{1.2.3.3.4.4.5.5.6.7}} = \frac{{2.6}}{7} = \frac{{12}}{7}\)