Giải bài 123 trang 60 sách bài tập Toán 6 – Cánh Diều Tập 2

Đề bài

Tính một cách hợp lí:

a) \(1,6 + \left( {2,7 – 0,7.6} \right) – \,\left( {94.0,7 – 99.2,7} \right);\)

b) \(0,1 – \,0,02 + 0,2 – 0,01 + 0,03 – 0,8\)

c) \(\left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right)\,.\,\left( {\frac{5}{6}\, – \frac{1}{2}\, – \,\frac{1}{3}} \right);\)

d) \(\left( {\frac{2}{{1.3}}\, + \,\frac{2}{{3.5}} + \frac{2}{{5.7}}} \right)\,.\,\left( {\frac{{10.13}}{3} – \frac{{{2^2}}}{3} – \frac{{{5^3}}}{3}} \right)\,\)

e) \(\frac{{2\;.\;4\;.\;10\; + \;4\;.\;6\;.\;8\; + \;14\;.\;16\;.\;20}}{{3\;.\;6\;.\;15\; + \;6\;.\;9\;.\;12\; + \;21\;.\;24\;.\;30}}\)

Lời giải chi tiết

a)

 \(\begin{array}{l}1,6 + \left( {2,7 – 0,7.6} \right) – \,\left( {94.0,7 – 99.2,7} \right)\\ = 1,6 + 2,7 – 0,7.6 – \,94.0,7 + 99.2,7\\ = 1,6 + \left( {2,7 + 99.2,7} \right) – \left( {0,7.6 + \,94.0,7} \right)\\ = 1,6 + 2,7.\left( {1 + 99} \right) – 0,7.\left( {6 + \,94} \right)\\ = 1,6 + 2,7.100 – 0,7.100\\ = 1,6 + 270 – 70\\ = 1,6 + \left( {270 – 70} \right)\\ = 1,6 + 200\\ = 201,6\end{array}\)

b)

 \(\begin{array}{l}0,1 – \,0,02 + 0,2 – 0,01 + 0,03 – 0,8\\ = \left( {0,1 + 0,2 – 0,8} \right) + \left( {0,03 – \,0,02 – 0,01} \right)\\ =  – 0,5 + 0\\ =  – 0,5\end{array}\)

c)

 \(\begin{array}{l}\left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right)\,.\,\left( {\frac{5}{6}\, – \frac{1}{2}\, – \,\frac{1}{3}} \right);\\ = \left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right)\,.\,\left( {\frac{5}{6}\, – \frac{{1.3}}{{2.3}}\, – \,\frac{{1.2}}{{3.2}}} \right)\\ = \left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right)\,.\,\left( {\frac{5}{6}\, – \frac{3}{6}\, – \,\frac{2}{6}} \right)\\ = \left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right)\,.\,\left( {\frac{{5 – 3 – 2}}{6}\,} \right)\\ = \left( {\frac{{ – 5}}{{116}} + \,\frac{{ – 117}}{{232}} – \frac{{71}}{{464}}} \right).0\\ = 0\end{array}\)

d)

 \(\begin{array}{l}\left( {\frac{2}{{1.3}}\, + \,\frac{2}{{3.5}} + \frac{2}{{5.7}}} \right)\,.\,\left( {\frac{{10.13}}{3} – \frac{{{2^2}}}{3} – \frac{{{5^3}}}{3}} \right)\,\\ = \left( {\frac{1}{1} – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + \frac{1}{5} – \frac{1}{7}} \right).\left( {\frac{{10.13 – {2^2} – {5^3}}}{3}} \right)\\ = \left( {1 – \frac{1}{7}} \right).\frac{{130 – 4 – 125}}{3}\\ = \frac{6}{7}.\frac{1}{3}\\ = \frac{2}{7}\end{array}\)

e)

 \(\begin{array}{l}\frac{{2\;.\;4\;.\;10\; + \;4\;.\;6\;.\;8\; + \;14\;.\;16\;.\;20}}{{3\;.\;6\;.\;15\; + \;6\;.\;9\;.\;12\; + \;21\;.\;24\;.\;30}}\\ = \frac{{2.2.2\left( {1\;.\;2\;.\;5\; + \;2\;.\;3\;.\;4\; + \;7\;.\;8\;.\;10} \right)}}{{3.3.3\left( {1\;.\;2\;.\;5\; + \;2\;.\;3\;.\;4\; + \;7\;.\;8\;.\;10} \right)}}\\ = \frac{8}{27}\end{array}\)