Giải bài 50 trang 42 sách bài tập Toán 6 – Cánh Diều Tập 2

Đề bài

Tính

a) \(\frac{{\frac{3}{5} + \frac{3}{{27}} – \frac{3}{9} – \frac{3}{{11}}}}{{\frac{4}{5} + \frac{4}{{27}} – \frac{4}{9} – \frac{4}{{11}}}}\)

b) \(\frac{{5 – \frac{5}{3} – \frac{5}{{27}}}}{{8 – \frac{8}{3} – \frac{8}{{27}}}}:\frac{{15 + \frac{{15}}{{121}} – \frac{{15}}{{11}}}}{{16 + \frac{{16}}{{121}} – \frac{{16}}{{11}}}}\)

c) \(\frac{1}{2}:\left( {{\kern 1pt} \frac{{ – 3}}{2}} \right):\frac{4}{3}:\left( {{\kern 1pt} \frac{{ – 5}}{4}} \right):\frac{6}{5}:\left( {{\kern 1pt} \frac{{ – 7}}{6}} \right):…:\left( {{\kern 1pt} \frac{{ – 101}}{{100}}} \right)\)

Lời giải chi tiết

a) \(\frac{{\frac{3}{5} + \frac{3}{{27}} – \frac{3}{9} – \frac{3}{{11}}}}{{\frac{4}{5} + \frac{4}{{27}} – \frac{4}{9} – \frac{4}{{11}}}} = \frac{{3.\left( {\frac{1}{5} + \frac{1}{{27}} – \frac{1}{9} – \frac{1}{{11}}} \right)}}{{4.\left( {\frac{1}{5} + \frac{1}{{27}} – \frac{1}{9} – \frac{1}{{11}}} \right)}} = \frac{3}{4}\)

b) \(\frac{{5 – \frac{5}{3} – \frac{5}{{27}}}}{{8 – \frac{8}{3} – \frac{8}{{27}}}}:\frac{{15 + \frac{{15}}{{121}} – \frac{{15}}{{11}}}}{{16 + \frac{{16}}{{121}} – \frac{{16}}{{11}}}} = \frac{{5.\left( {1 – \frac{1}{3} – \frac{1}{{27}}} \right)}}{{8.\left( {1 – \frac{1}{3} – \frac{1}{{27}}} \right)}}:\frac{{15.\left( {1 + \frac{1}{{121}} – \frac{1}{{11}}} \right)}}{{16.\left( {1 + \frac{1}{{121}} – \frac{1}{{11}}} \right)}} = \frac{5}{8}:\frac{{15}}{{16}} = \frac{5}{8}.\frac{{16}}{{15}} = \frac{2}{3}\)

c)

\(\begin{array}{l}\frac{1}{2}:\left( {{\kern 1pt} \frac{{ – 3}}{2}} \right):\frac{4}{3}:\left( {{\kern 1pt} \frac{{ – 5}}{4}} \right):\frac{6}{5}:\left( {{\kern 1pt} \frac{{ – 7}}{6}} \right):…:\left( {{\kern 1pt} \frac{{ – 101}}{{100}}} \right) = \frac{1}{2}.\left( {{\kern 1pt} \frac{2}{{ – 3}}} \right).\frac{3}{4}.\left( {{\kern 1pt} \frac{4}{{ – 5}}} \right).\frac{5}{6}.\left( {{\kern 1pt} \frac{6}{{ – 7}}} \right)…\left( {{\kern 1pt} \frac{{100}}{{ – 101}}} \right)\\ = \frac{{1.2.3.4.5.6….100}}{{2.( – 3).4.( – 5).6.( – 7)…( – 101)}} = \frac{{1.2.3.4.5.6….100}}{{2.( – 3).4.( – 5).6.( – 7)…( – 101)}} = \frac{{1.2.3.4.5.6…100}}{{2.3.4.5.6.7…101}} = \frac{1}{{101}}\end{array}\)

Vì có chẵn số âm nên khi lấy tích ta bỏ các dấu âm đi.