Bài 122 trang 59 sách bài tập Toán 6 – Cánh Diều Tập 2

Đề bài

Thực hiện phép tính:

a) \(0,58\,.\,{7^2} - \left( { - 7} \right)\,.\,\left( { - 0,7} \right)\,.\,15,8;\)

b) \(0,05\,:\,0,5 + 7\,:\,0,7 + 0,9:0,009;\)

c) \(\frac{9}{{11}}\,.\,\frac{{92}}{{121}} + \frac{2}{{ - 121}}\,.\,\frac{9}{{11}}\, + \frac{{31}}{{121}}\,.\,\frac{9}{{11}};\)

d) \(\frac{{20\,212\,021}}{{2\,021}}\,.\,\frac{{2\,020}}{{20\,202\,020}}.\frac{{{2^3}}}{{{3^2}}}\,.\,\frac{{ - 3}}{{{2^2}}}\,\)

Lời giải chi tiết

a)

\(\begin{array}{l}0,58\,.\,{7^2} - \left( { - 7} \right)\,.\,\left( { - 0,7} \right)\,.\,15,8\\ = 0,58\,.\,{7^2} - 7\,.\,0,7\,.\,15,8\\ = 0,58\,.\,{7^2} - 7\,.\,7\,.\,1,58\\ = {7^2}\left( {0,58 - 1,58} \right)\\ = {7^2}.( - 1)\\ =  - 49\end{array}\)

b)

 \(\begin{array}{l}0,05\,:\,0,5 + 7\,:\,0,7 + 0,9:0,009\\ = \frac{5}{{100}}:\frac{5}{{10}} + 7:\frac{7}{{10}} + \frac{9}{{10}}:\frac{9}{{1000}}\\ = \frac{5}{{100}}.\frac{{10}}{5} + 7.\frac{{10}}{7} + \frac{9}{{10}}.\frac{{1000}}{9}\\ = 0,1 + 10 + 100\\ = 110,1.\end{array}\)

c)

 \(\begin{array}{l}\frac{9}{{11}}\,.\,\frac{{92}}{{121}} + \frac{2}{{ - 121}}\,.\,\frac{9}{{11}}\, + \frac{{31}}{{121}}\,.\,\frac{9}{{11}}\\ = \frac{9}{{11}}\,.\,\left( {\frac{{92}}{{121}} + \frac{2}{{ - 121}}\,\, + \frac{{31}}{{121}}} \right)\\ = \frac{9}{{11}}\,.\,\left( {\frac{{92}}{{121}} + \frac{{ - 2}}{{121}}\,\, + \frac{{31}}{{121}}} \right)\\ = \frac{9}{{11}}\,.\,\frac{{92 + ( - 2) + 31}}{{121}}\\ = \frac{9}{{11}}\,.\,\frac{{121}}{{121}}\\ = \frac{9}{{11}}.\end{array}\)

d)

 \(\begin{array}{l}\frac{{20\,212\,021}}{{2\,021}}\,.\,\frac{{2\,020}}{{20\,202\,020}}.\frac{{{2^3}}}{{{3^2}}}\,.\,\frac{{ - 3}}{{{2^2}}}\,\\ = \left( {\frac{{20\,212\,021}}{{2\,021}}\,.\,\frac{{2\,020}}{{20\,202\,020}}} \right).\left( {\frac{{{2^3}}}{{{3^2}}}\,.\,\frac{{ - 3}}{{{2^2}}}} \right)\,\\ = \left( {\frac{{2021.10\,001}}{{2\,021}}\,.\,\frac{{2\,020}}{{2020.10\,001}}} \right).\left( {\frac{{{2^3}.3.( - 1)}}{{{3^2}{{.2}^2}}}\,} \right)\,\\ = \frac{{2\,021.10\,001.\,2\,020}}{{2\,021.\,2\,020\,.10\,001}}\,\,.\,\,\frac{{2.( - 1)}}{3}\\ = 1.\frac{{( - 2)}}{3}\\ = \frac{{ - 2}}{3}\end{array}\)