Thực hiện phép tính \(\left( {\dfrac{{ - 1}}{3} + \dfrac{3}{{11}}} \right).\dfrac{2}{{2021}} + \left( {\dfrac{1}{3} + \dfrac{8}{{11}}} \right).\dfrac{2}{{2021}}\) ta được kết quả là:
\(\left( {\dfrac{{ - 1}}{3} + \dfrac{3}{{11}}} \right).\dfrac{2}{{2021}} + \left( {\dfrac{1}{3} + \dfrac{8}{{11}}} \right).\dfrac{2}{{2021}}\)
\( = \left( {\dfrac{{ - 1}}{3} + \dfrac{3}{{11}} + \dfrac{1}{3} + \dfrac{8}{{11}}} \right).\dfrac{2}{{2021}}\)
\( = \left[ {\left( {\dfrac{{ - 1}}{3} + \dfrac{1}{3}} \right) + \left( {\dfrac{3}{{11}} + \dfrac{8}{{11}}} \right)} \right].\dfrac{2}{{2021}}\)
\( = \left( {0 + 1} \right).\dfrac{2}{{2021}}\)
\( = \dfrac{2}{{2021}}\).
Phép tính \(\dfrac{3}{5}.\left( {\dfrac{5}{3} - \dfrac{2}{7}} \right) - \left( {\dfrac{7}{3} - \dfrac{9}{7}} \right).\dfrac{3}{5}\) có kết quả là:
\(\dfrac{3}{5}.\left( {\dfrac{5}{3} - \dfrac{2}{7}} \right) - \left( {\dfrac{7}{3} - \dfrac{9}{7}} \right).\dfrac{3}{5}\)
\( = \dfrac{3}{5}.\left[ {\left( {\dfrac{5}{3} - \dfrac{2}{7}} \right) - \left( {\dfrac{7}{3} - \dfrac{9}{7}} \right)} \right]\)
\( = \dfrac{3}{5}.\left( {\dfrac{5}{3} - \dfrac{2}{7} - \dfrac{7}{3} + \dfrac{9}{7}} \right)\)
\( = \dfrac{3}{5}.\left[ {\left( {\dfrac{5}{3} - \dfrac{7}{3}} \right) + \left( {\dfrac{{ - 2}}{7} + \dfrac{9}{7}} \right)} \right]\)
\( = \dfrac{3}{5}.\left( {\dfrac{{ - 2}}{3} + 1} \right)\)
\( = \dfrac{3}{5}.\dfrac{1}{3} = \dfrac{1}{5}\).
Giá trị \(x\) thỏa mãn \(\dfrac{5}{6}.x = \dfrac{{ - 3}}{5}\) là:
\(\begin{array}{l}\dfrac{5}{6}.x = \dfrac{{ - 3}}{5}\\x = \dfrac{{ - 3}}{5}:\dfrac{5}{6}\\x = \dfrac{{ - 3}}{5}.\dfrac{6}{5}\\x = \dfrac{{ - 18}}{{25}}\end{array}\).
Vậy \(x = \dfrac{-18}{{25}}.\)
Tìm x, biết \(\dfrac{6}{5}.x = \dfrac{{ - 2}}{{15}}\):
\(\begin{array}{l}\dfrac{6}{5}.x = \dfrac{{ - 2}}{{15}}\\x = \dfrac{{ - 2}}{{15}}:\dfrac{6}{5}\\x = \dfrac{{ - 2}}{{15}}.\dfrac{5}{6}\\x = \dfrac{{ - 2.5}}{{3.5.2.3}}\\x = \dfrac{{ - 1}}{9}\end{array}\).
Vậy \(x = \dfrac{-1}{{9}}.\)
Tích \(\dfrac{{ - 7}}{{15}}.\dfrac{5}{{ - 21}}\) là:
\(\dfrac{{ - 7}}{{15}}.\dfrac{5}{{ - 21}} = \dfrac{{( - 7).5}}{{15.21}} = \dfrac{{\left( { - 7} \right).5}}{{3.5.3.( - 7)}} = \dfrac{1}{9}\).
Ta thấy \(\dfrac{1}{9}\) là số hữu tỉ dương.
Kết quả của phép tính \(\dfrac{{ - 1}}{3}.\dfrac{{ - 2}}{5}\) là:
Thực hiện phép nhân \(\dfrac{{ - 8}}{5}.\dfrac{{ - 3}}{4}\) ta được:
\(\dfrac{{ - 8}}{5}.\dfrac{{ - 3}}{4} = \dfrac{{( - 8).( - 3)}}{{5.4}} = \dfrac{{24}}{{20}} = \dfrac{6}{5}\).
Giá trị biểu thức \(\dfrac{1}{2} - 2.\left[ {\dfrac{1}{3} - \left( {\dfrac{{ - 5}}{4} + \dfrac{1}{3}} \right)} \right] + 0,25\) là:
\(\begin{array}{l}\dfrac{1}{2} - 2.\left[ {\dfrac{1}{3} - \left( {\dfrac{{ - 5}}{4} + \dfrac{1}{3}} \right)} \right] + 0,25\\ = \dfrac{1}{2} - 2.\left[ {\dfrac{1}{3} - \left( {\dfrac{{ - 15}}{{12}} + \dfrac{4}{{12}}} \right)} \right] + \dfrac{1}{4}\\ = \dfrac{1}{2} - 2.\left[ {\dfrac{1}{3} - \left( {\dfrac{{ - 11}}{{12}}} \right)} \right] + \dfrac{1}{4}\\ = \dfrac{1}{2} - 2.\left[ {\dfrac{4}{{12}} + \dfrac{{11}}{{12}}} \right] + \dfrac{1}{4}\\ = \dfrac{1}{2} - 2.\dfrac{{15}}{{12}} + \dfrac{1}{4}\\ = \dfrac{1}{2} - \dfrac{{30}}{{12}} + \dfrac{1}{4}\\ = \dfrac{6}{{12}} - \dfrac{{30}}{{12}} + \dfrac{3}{{12}}\\ = \dfrac{{ - 21}}{{12}} = \dfrac{{ - 7}}{4}\end{array}\)
Giá trị biểu thức \(\dfrac{{31}}{4}:\left[ {\dfrac{1}{7} - \left( {\dfrac{3}{4} + \dfrac{1}{2}} \right)} \right] + 0,5\) bằng:
\(\begin{array}{l}\dfrac{{31}}{4}:\left[ {\dfrac{1}{7} - \left( {\dfrac{3}{4} + \dfrac{1}{2}} \right)} \right] + 0,5\\ = \dfrac{{31}}{4}:\left[ {\dfrac{1}{7} - \left( {\dfrac{3}{4} + \dfrac{2}{4}} \right)} \right] + \dfrac{1}{2}\\ = \dfrac{{31}}{4}:\left( {\dfrac{1}{7} - \dfrac{5}{4}} \right) + \dfrac{1}{2}\\ = \dfrac{{31}}{4}:\left( {\dfrac{4}{{28}} - \dfrac{{35}}{{28}}} \right) + \dfrac{1}{2}\\ = \dfrac{{31}}{4}:\dfrac{{( - 31)}}{{28}} + \dfrac{1}{2}\\ = \dfrac{{31}}{4}.\dfrac{{( - 28)}}{{31}} + \dfrac{1}{2}\\ = \dfrac{{ - 28}}{4} + \dfrac{1}{2}\\ = \dfrac{{ - 14}}{2} + \dfrac{1}{2}\\ = \dfrac{{ - 13}}{2}.\end{array}\)
Tính giá trị biểu thức: \(3,5 - \left[ {\dfrac{1}{3}:\left( {1 + \dfrac{{11}}{3}} \right)} \right].\dfrac{7}{4}\) ta được kết quả là:
\(\begin{array}{l}3,5 - \left[ {\dfrac{1}{3}:\left( {1 + \dfrac{{11}}{3}} \right)} \right].\dfrac{7}{4}\\ = \dfrac{7}{2} - \left[ {\dfrac{1}{3}:\left( {\dfrac{3}{3} + \dfrac{{11}}{3}} \right)} \right].\dfrac{7}{4}\\ = \dfrac{7}{2} - \left[ {\dfrac{1}{3}:\dfrac{{14}}{3}} \right].\dfrac{7}{4}\\ = \dfrac{7}{2} - \left[ {\dfrac{1}{3}.\dfrac{3}{{14}}} \right].\dfrac{7}{4}\\ = \dfrac{7}{2} - \dfrac{1}{{14}}.\dfrac{7}{4}\\ = \dfrac{7}{2} - \dfrac{{1.7}}{{2.7.4}}\\ = \dfrac{7}{2} - \dfrac{1}{8}\\ = \dfrac{{28}}{8} - \dfrac{1}{8}\\ = \dfrac{{27}}{8}.\end{array}\)
Ta có \(\left( {\dfrac{2}{3}x - \dfrac{4}{9}} \right)\left( {\dfrac{1}{2} + \dfrac{{ - 3}}{7}:x} \right) = 0\,\)
TH1: \(\dfrac{2}{3}x - \dfrac{4}{9} = 0\)
\(\dfrac{2}{3}x = \dfrac{4}{9}\)
\(x = \dfrac{4}{9}:\dfrac{2}{3}\)
\(x = \dfrac{4}{9}.\dfrac{3}{2}\)
\(x = \dfrac{2}{3}\)
TH2: \(\dfrac{1}{2} + \dfrac{{ - 3}}{7}:x = 0\)
\(\dfrac{{ - 3}}{7}:x = \dfrac{{ - 1}}{2}\)
\(x = \dfrac{{ - 3}}{7}:\left( {\dfrac{{ - 1}}{2}} \right)\)
\(x = \dfrac{6}{7}\)
Vậy có hai giá trị của \(x\) thỏa mãn là \(x = \dfrac{2}{3};x = \dfrac{6}{7}\) .
Ta có: \(\left( {x:\dfrac{2}{5} + \dfrac{1}{6}} \right)\left( {\dfrac{{14}}{{15}} + \dfrac{1}{5}.x} \right) = 0\,\)
Suy ra \(x:\dfrac{2}{5} + \dfrac{1}{6} = 0\,\) hoặc \(\dfrac{{14}}{{15}} + \dfrac{1}{5}.x = 0\,\)
TH1: \(x:\dfrac{2}{5} + \dfrac{1}{6} = 0\,\)
\(x:\dfrac{2}{5} = \dfrac{{ - 1}}{6}\)
\(x = \dfrac{{ - 1}}{6}.\dfrac{2}{5}\)
\(x = \dfrac{{ - 1.2}}{{6.5}}\)
\(x = \dfrac{{ - 1.2}}{{2.3.5}}\)
\(x = \dfrac{{ - 1}}{{15}}\)
TH2: \(\dfrac{{14}}{{15}} + \dfrac{1}{5}.x = 0\,\)
\(\dfrac{1}{5}.x = \dfrac{{ - 14}}{{15}}\)
\(x = \dfrac{{ - 14}}{{15}}:\dfrac{1}{5}\)
\(x = \dfrac{{ - 14}}{{15}}.\dfrac{5}{1}\)
\(x = \dfrac{{ - 14.5}}{{3.5}}\)
\(x = \dfrac{{ - 14}}{3}\)
Do đó có hai giá trị của \(x\) thỏa mãn \(\left( {x:\dfrac{2}{5} + \dfrac{1}{6}} \right)\left( {\dfrac{{14}}{{15}} + \dfrac{1}{5}.x} \right) = 0\,\) là \(x = \dfrac{{ - 1}}{{15}}\); \(x = \dfrac{{ - 14}}{3}\).
Nếu \(x = \dfrac{a}{b};\,y = \dfrac{c}{d}\,\left( {b,d \ne 0} \right)\) thì tích \(x.y\) bằng
Với \(x = \dfrac{a}{b};\,y = \dfrac{c}{d}\,\left( {b,d \ne 0} \right)\) ta có: \(x.y = \dfrac{a}{b}.\dfrac{c}{d} = \dfrac{{a.c}}{{b.d}}\) .
Kết quả của phép tính \( - \dfrac{6}{7}.\dfrac{{21}}{{12}}\) là
Ta có \( - \dfrac{6}{7}.\dfrac{{21}}{{12}} = - \dfrac{6}{7}.\dfrac{7}{4} = \dfrac{{ - 6}}{4} = - \dfrac{3}{2}\)
Thực hiện phép tính $\dfrac{5}{{11}}:\dfrac{{15}}{{22}}$ ta được kết quả là:
Ta có $\dfrac{5}{{11}}:\dfrac{{15}}{{22}}$\( = \dfrac{5}{{11}}.\dfrac{{22}}{{15}} = \dfrac{{5.22}}{{11.15}} = \dfrac{2}{3}\)
Kết quả của phép tính $\dfrac{3}{2}.\dfrac{4}{7}$ là
Ta có $\dfrac{3}{2}.\dfrac{4}{7} = \dfrac{{3.4}}{{2.7}} = \dfrac{6}{7} > 0$
Số nào sau đây là kết quả của phép tính \(1\dfrac{4}{5}:\left( { - \dfrac{3}{4}} \right)\)
Ta có \(1\dfrac{4}{5}:\left( { - \dfrac{3}{4}} \right)\)\( = \dfrac{9}{5}.\left( { - \dfrac{4}{3}} \right) = - \dfrac{{9.4}}{{5.3}} = - \dfrac{{12}}{5}\)
Cho \(A = \dfrac{{ - 5}}{6}.\dfrac{{12}}{{ - 7}}.\left( {\dfrac{{ - 21}}{{15}}} \right);\,B = \dfrac{1}{6}.\dfrac{9}{{ - 8}}.\left( {\dfrac{{ - 12}}{{11}}} \right)\) . So sánh \(A\) và \(B\).
Ta có
\(A = \dfrac{{ - 5}}{6}.\dfrac{{12}}{{ - 7}}.\left( {\dfrac{{ - 21}}{{15}}} \right) = \dfrac{{\left( { - 5} \right).12.\left( { - 21} \right)}}{{6.\left( { - 7} \right).15}} = \dfrac{{\left( { - 5} \right).2.6.\left( { - 7} \right).3}}{{6.\left( { - 7} \right).5.3}} = - 2\)
\(B = \dfrac{1}{6}.\dfrac{9}{{ - 8}}.\left( {\dfrac{{ - 12}}{{11}}} \right) = \dfrac{{9.\left( { - 12} \right)}}{{6.\left( { - 8} \right).11}} = \dfrac{9}{{44}}\)
Suy ra \(A < B\) .
Tìm \(x\) biết \(\dfrac{2}{3}x = - \dfrac{1}{{8}}.\)
Ta có \(\dfrac{2}{3}x = - \dfrac{1}{{8}}\)
\(x = \left( { - \dfrac{1}{{8}}} \right):\dfrac{2}{3}\)
\(x = \dfrac{{ - 1}}{8}.\dfrac{3}{2}\)
\(x = - \dfrac{3}{{16}}\)
Vậy \(x = - \dfrac{3}{{16}}.\)
Tìm số $x$ thoả mãn: \(x:\left( {\dfrac{2}{5} - 1\dfrac{2}{5}} \right) = 1.\)
Ta có \(x:\left( {\dfrac{2}{5} - 1\dfrac{2}{5}} \right) = 1\)
\(x:\left( {\dfrac{2}{5} - \dfrac{7}{5}} \right) = 1\)
\(x:\left( {\dfrac{{ - 5}}{5}} \right) = 1\)
\(x:\left( { - 1} \right) = 1\)
\(x = 1.\left( { - 1} \right)\)
\(x = - 1\)
Vậy \(x = - 1\) .
