$A = \dfrac{{\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}}}{{\dfrac{8}{3} - \dfrac{8}{5} + \dfrac{8}{{10}}}} + \dfrac{1}{2}$
$A = \dfrac{{\left( {\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}} \right)}}{{4.\left( {\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}} \right)}} + \dfrac{1}{2}$
$A = \dfrac{1}{4} + \dfrac{1}{2}$
$A = \dfrac{3}{4}.$
Ta có: $A = \dfrac{3}{4} < \dfrac{4}{4} =1.$
Vậy \( A<1\)
\(A = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{{15}}{{119}}}}{{\dfrac{{ - 10}}{{68}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{{15}}{{238}}}}\)
\( = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{{3.5}}{{7.17}}}}{{\dfrac{{ - 2.5}}{{4.17}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{{3.5}}{{14.17}}}}\)
\( = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{3}{7}.\dfrac{5}{{17}}}}{{\dfrac{{ - 1}}{2}.\dfrac{5}{{17}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{3}{{14}}.\dfrac{5}{{17}}}}\)
\( = \dfrac{{\dfrac{5}{{17}}.\left( {\dfrac{1}{2} - \dfrac{{13}}{{14}} + \dfrac{3}{7}} \right)}}{{\dfrac{5}{{17}}.\left( {\dfrac{{ - 1}}{2} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}} \right)}}\)
\( = \dfrac{{\dfrac{1}{2} - \dfrac{{13}}{{14}} + \dfrac{3}{7}}}{{\dfrac{{ - 1}}{2} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}}}\)
\( = \dfrac{{\dfrac{7}{{14}} - \dfrac{{13}}{{14}} + \dfrac{6}{{14}}}}{{\dfrac{{ - 7}}{{14}} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}}}\)
\( = \dfrac{{\dfrac{0}{{14}}}}{{\dfrac{{16}}{{14}}}}\)
\( = 0\)
Giá trị \(x\) thỏa mãn \(\dfrac{{11}}{3}:\left( {\dfrac{1}{2}x - \dfrac{5}{3}} \right) + 1 = \dfrac{1}{9}\) là:
\(\begin{array}{l}\dfrac{{11}}{3}:\left( {\dfrac{1}{2}x - \dfrac{5}{3}} \right) + 1 = \dfrac{1}{9}\\\dfrac{{11}}{3}:\left( {\dfrac{1}{2}x - \dfrac{5}{3}} \right) = \dfrac{1}{9} - 1\\\dfrac{{11}}{3}:\left( {\dfrac{1}{2}x - \dfrac{5}{3}} \right) = \dfrac{{ - 8}}{9}\\\dfrac{1}{2}x - \dfrac{5}{3} = \dfrac{{11}}{3}:\dfrac{{ - 8}}{9}\\\dfrac{1}{2}x - \dfrac{5}{3} = \dfrac{{11}}{3}.\dfrac{{ - 9}}{8}\\\dfrac{1}{2}x - \dfrac{5}{3} = \dfrac{{ - 33}}{8}\\\dfrac{1}{2}x = \dfrac{{ - 33}}{8} + \dfrac{5}{3}\\\dfrac{1}{2}x = \dfrac{{ - 99}}{{24}} + \dfrac{{40}}{{24}}\\\dfrac{1}{2}x = \dfrac{{ - 59}}{{24}}\\x = \dfrac{{ - 59}}{{24}}:\dfrac{1}{2}\\x = \dfrac{{ - 59}}{{24}}.2\\x = \dfrac{{ - 59}}{{12}}.\end{array}\).
Vậy \(x = \dfrac{{ - 59}}{{12}}\).
Ta có: \(\dfrac{3}{7} + \dfrac{1}{7}:x = \dfrac{3}{{14}}\)
\(\dfrac{1}{7}:x = \dfrac{3}{{14}} - \dfrac{3}{7}\)
\(\dfrac{1}{7}:x = \dfrac{3}{{14}} - \dfrac{6}{{14}}\)
\(\dfrac{1}{7}:x = \dfrac{{ - 3}}{{14}}\)
\(x = \dfrac{1}{7}:\left( {\dfrac{{ - 3}}{{14}}} \right)\)
\(x = \dfrac{1}{7}.\dfrac{{14}}{{\left( { - 3} \right)}}\)
\(x = - \dfrac{2}{3}\)
Vậy \({x_1} = - \dfrac{2}{3}\)
* \(\dfrac{5}{7} + \dfrac{2}{7}:x = 1\)
\(\dfrac{2}{7}:x = 1 - \dfrac{5}{7}\)
\(\dfrac{2}{7}:x = \dfrac{2}{7}\)
\(x = \dfrac{2}{7}:\dfrac{2}{7}\)
\(x = 1\)
Vậy \({x_2} = 1\) .
Do đó:
\({x_1} + {x_2} = \dfrac{{ - 2}}{3} + 1 = \dfrac{{ - 2}}{3} + \dfrac{3}{3} = \dfrac{1}{3}\).
+) \(x:\left( { - 2\dfrac{1}{{15}}} \right) + 3\dfrac{1}{2} = - \dfrac{3}{4}\)
\(x:\dfrac{{ - 31}}{{15}} + \dfrac{7}{2} = - \dfrac{3}{4}\)
\(x:\dfrac{{ - 31}}{{15}} = - \dfrac{3}{4} - \dfrac{7}{2}\)
\(x:\dfrac{{ - 31}}{{15}} = \dfrac{{ - 3}}{4} - \dfrac{{14}}{4}\)
\(x:\dfrac{{ - 31}}{{15}} = \dfrac{{ - 17}}{4}\)
\(x = \dfrac{{ - 17}}{4}.\dfrac{{ - 31}}{{15}}\)
\(x = \dfrac{{ - 17.( - 31)}}{{4.15}}\)
\(x = \dfrac{{527}}{{60}}\)
Vậy \({x_1} = \dfrac{{527}}{{60}}\)
+) \(\dfrac{5}{{11}} + \dfrac{6}{{11}}:x = 2\)
\(\dfrac{6}{{11}}:x = 2 - \dfrac{5}{{11}}\)
\(\dfrac{6}{{11}}:x = \dfrac{{22}}{{11}} - \dfrac{5}{{11}}\)
\(\dfrac{6}{{11}}:x = \dfrac{{17}}{{11}}\)
\(x = \dfrac{6}{{11}}:\dfrac{{17}}{{11}}\)
\(x = \dfrac{6}{{11}}.\dfrac{{11}}{{17}}\)
\(x = \dfrac{6}{{17}}\)
Vậy \({x_2} = \dfrac{6}{{17}}\)
Do đó:
\({x_1} + {x_2} = \dfrac{{527}}{{60}} + \dfrac{6}{{17}} = \dfrac{{8959}}{{1020}} + \dfrac{{360}}{{1020}} = \dfrac{{9319}}{{1020}}\).
Ta có \(\dfrac{1}{3}x + \dfrac{2}{5}\left( {x - 1} \right) = 0\)
\(\dfrac{1}{3}x + \dfrac{2}{5}x - \dfrac{2}{5} = 0\)
\(\dfrac{1}{3}x + \dfrac{2}{5}x = \dfrac{2}{5}\)
\(x\left( {\dfrac{1}{3} + \dfrac{2}{5}} \right) = \dfrac{2}{5}\)
\(x.\left( {\dfrac{5}{{15}} + \dfrac{6}{{15}}} \right) = \dfrac{2}{5}\)
\(x.\dfrac{{11}}{{15}} = \dfrac{2}{5}\)
\(x = \dfrac{2}{5}:\dfrac{{11}}{{15}}\)
\(x = \dfrac{2}{5}.\dfrac{{15}}{{11}}\)
\(x = \dfrac{{2.15}}{{5.11}}\)
\(x = \dfrac{6}{{11}}\)
Vậy \(x = \dfrac{6}{{11}}\) thoả mãn.
Ta có: \(\left( {3\dfrac{5}{7}x - 1\dfrac{5}{7}x} \right) - \dfrac{1}{3} = \dfrac{2}{3}\)
\(\left( {\dfrac{{26}}{7}x - \dfrac{{12}}{7}x} \right) - \dfrac{1}{3} = \dfrac{2}{3}\)
\(\left( {\dfrac{{26}}{7} - \dfrac{{12}}{7}} \right)x - \dfrac{1}{3} = \dfrac{2}{3}\)
\(\dfrac{{14}}{7}x - \dfrac{1}{3} = \dfrac{2}{3}\)
\(2x - \dfrac{1}{3} = \dfrac{2}{3}\)
\(2x = \dfrac{2}{3} + \dfrac{1}{3}\)
\(2x = 1\)
\(x = \dfrac{1}{2}\)
Vậy \(x = \dfrac{1}{2}\) thỏa mãn.
Gọi \({x_0}\) là giá trị thỏa mãn \(\dfrac{{-1}}{{12}}:x-\dfrac{5}{4}=-1\dfrac{1}{3}\), kết luận nào sau đây đúng:
\(\begin{array}{l}\dfrac{{-1}}{{12}}:x-\dfrac{5}{4}=-1\dfrac{1}{3}\\\dfrac{{-1}}{{12}}:x-\dfrac{5}{4}=-\dfrac{4}{3}\\\dfrac{{-1}}{{12}}:x=-\dfrac{4}{3}+\dfrac{5}{4}\\\dfrac{{-1}}{{12}}:x=-\dfrac{{16}}{{12}}+\dfrac{{15}}{{12}}\\\dfrac{{-1}}{{12}}:x=\dfrac{{-1}}{{12}}\\x=\dfrac{{-1}}{{12}}:\dfrac{{-1}}{{12}}\\x=1\end{array}\).
Vậy \(x_0=1\).
Ta có
\(A = \dfrac{{ - 5}}{6}.\dfrac{{12}}{{ - 7}}.\left( {\dfrac{{ - 21}}{{15}}} \right) = \dfrac{{\left( { - 5} \right).12.\left( { - 21} \right)}}{{6.\left( { - 7} \right).15}} = \dfrac{{\left( { - 5} \right).2.6.\left( { - 7} \right).3}}{{6.\left( { - 7} \right).5.3}} = - 2\)
\(B = \dfrac{1}{6}.\dfrac{9}{{ - 8}}.\left( {\dfrac{{ - 12}}{{11}}} \right) = \dfrac{{9.\left( { - 12} \right)}}{{6.\left( { - 8} \right).11}} = \dfrac{9}{{44}}\)
Do đó
\(A+B=\left({-2}\right)+\dfrac{9}{{44}}=\left({\dfrac{{-88}}{{44}}}\right)+\dfrac{9}{{44}}=\dfrac{{-79}}{{44}}.\)
Điền cụm từ thích hợp vào chỗ trống: “Muốn nhân hai phân số với nhau thì ta …"
Muốn nhân hai phân số, ta nhân các tử với nhau và nhân các mẫu với nhau.
\(A = \dfrac{4}{5}.\dfrac{{20}}{8}.\left( { - \dfrac{4}{3}} \right) = \dfrac{{4.20.( - 4)}}{{5.8.3}} = \dfrac{{4.5.2.2.( - 4)}}{{5.2.4.3}} = \dfrac{{2.( - 4)}}{3} = \dfrac{{ - 8}}{3}\)
\(B = \dfrac{{ - 2}}{{11}}.\dfrac{5}{{18}}.\left( {\dfrac{{ - 121}}{{25}}} \right) = \dfrac{{ - 2.5.( - 121)}}{{11.18.25}} = \dfrac{{ - 2.5.11.( - 11)}}{{11.2.9.5.5}} = \dfrac{{( - 1).( - 11)}}{{9.5}} = \dfrac{{11}}{{45}}\)
Do đó:
\(A+B=-\dfrac{8}{3}+\dfrac{{11}}{{45}}=-\dfrac{{120}}{{45}}+\dfrac{{11}}{{45}}=\dfrac{{-109}}{{45}}\).
Kết quả của phép tính \(\frac{7}{4}.\left( { - \frac{2}{5}} \right)\) là:
Ta có: \(\dfrac{7}{4}.\left( { - \dfrac{2}{5}} \right) = \dfrac{7}{4}.\dfrac{{ - 2}}{5} = \dfrac{{7.( - 2)}}{{4.5}} = \dfrac{{7.( - 2)}}{{2.2.5}} = \dfrac{{7.( - 1)}}{{1.2.5}} = \dfrac{{ - 7}}{{10}}\).
Thực hiện phép tính \(\dfrac{{ - 7}}{3}:\dfrac{{14}}{9}\) ta được kết quả là:
Ta có: \(\dfrac{{ - 7}}{3}:\dfrac{{14}}{9}\) \( = \dfrac{{ - 7}}{3}.\dfrac{9}{{14}} = \dfrac{{ - 7.9}}{{3.14}} = \dfrac{{ - 7.3.3}}{{3.2.7}} = \dfrac{{ - 3}}{2}\).
Kết quả của phép tính \(0,7.\dfrac{4}{5}\) là:
Ta có: \(0,7.\dfrac{4}{5} = \dfrac{7}{{10}}.\dfrac{4}{5} = \dfrac{{7.4}}{{10.5}} = \dfrac{{7.2.2}}{{2.5.5}} = \dfrac{{7.2}}{{5.5}} = \dfrac{{14}}{{25}} > 0\).
Số nào sau đây là kết quả của phép tính \( - 2:\left( {1\dfrac{1}{6}} \right)\).
Ta có: \( - 2:\left( {1\dfrac{1}{6}} \right)\)\( = - 2:\dfrac{7}{6} = - 2.\dfrac{6}{7} = \dfrac{{ - 2.6}}{7} = \dfrac{{ - 12}}{7}\).
\(2\dfrac{1}{3}:\left({-\dfrac{6}{5}}\right)=\dfrac{7}{3}:\left({\dfrac{{-6}}{5}}\right)=\dfrac{7}{3}.\left({\dfrac{{-5}}{6}}\right)=\dfrac{{-35}}{{18}}\).
Cho \(A = \dfrac{4}{5}.\dfrac{{20}}{8}.\left( { - \dfrac{4}{3}} \right);\,B = \dfrac{{ - 2}}{{11}}.\dfrac{5}{{18}}.\left( {\dfrac{{ - 121}}{{25}}} \right)\). So sánh \(A\) và \(B\).
\(A = \dfrac{4}{5}.\dfrac{{20}}{8}.\left( { - \dfrac{4}{3}} \right) = \dfrac{{4.20.( - 4)}}{{5.8.3}} = \dfrac{{4.5.2.2.( - 4)}}{{5.2.4.3}} = \dfrac{{2.( - 4)}}{3} = \dfrac{{ - 8}}{3}\)
\(B = \dfrac{{ - 2}}{{11}}.\dfrac{5}{{18}}.\left( {\dfrac{{ - 121}}{{25}}} \right) = \dfrac{{ - 2.5.( - 121)}}{{11.18.25}} = \dfrac{{ - 2.5.11.( - 11)}}{{11.2.9.5.5}} = \dfrac{{( - 1).( - 11)}}{{9.5}} = \dfrac{{11}}{{45}}\)
Ta có: \(\dfrac{{ - 8}}{3} < 0\) (vì tử và mẫu khác dấu);
\(\dfrac{{11}}{{45}} > 0\) (vì tử và mẫu cùng dấu)
Do đó \(\dfrac{{ - 8}}{3} < \dfrac{{11}}{{45}}\) hay \(A < B\).
Tìm \(x\) biết \(\dfrac{5}{{11}}x = \dfrac{{25}}{{44}}.\)
Ta có: \(\dfrac{5}{{11}}x = \dfrac{{25}}{{44}}\)
\(x = \dfrac{{25}}{{44}}:\dfrac{5}{{11}}\)
\(x = \dfrac{{25}}{{44}}.\dfrac{{11}}{5}\)
\(x = \dfrac{{25.11}}{{44.5}}\)
\(x = \dfrac{{5.5.11}}{{4.11.5}}\)
\(x = \dfrac{5}{4}\)
Vậy \(x = \dfrac{5}{4}\).
Tìm số \(x\) thoả mãn: \(x:\left( {\dfrac{2}{9} - \dfrac{1}{5}} \right) = \dfrac{8}{{16}}.\)
Ta có: \(x:\left( {\dfrac{2}{9} - \dfrac{1}{5}} \right) = \dfrac{8}{{16}}\)
\(x:\left( {\dfrac{{10}}{{45}} - \dfrac{9}{{45}}} \right) = \dfrac{1}{2}\)
\(x:\dfrac{1}{{45}} = \dfrac{1}{2}\)
\(x = \dfrac{1}{2}.\dfrac{1}{{45}}\)
\(x = \dfrac{{1.1}}{{2.45}}\)
\(x = \dfrac{1}{{90}}\)
Vậy \(x = \dfrac{1}{{90}}\).
Gọi \({x_0}\) là giá trị thỏa mãn \(\left( {\dfrac{6}{7} - \dfrac{3}{5}} \right):\dfrac{{2x}}{3} = - \dfrac{{11}}{{18}}\). Chọn câu đúng.
Ta có:
\(\left( {\dfrac{6}{7} - \dfrac{3}{5}} \right):\dfrac{{2x}}{3} = - \dfrac{{11}}{{18}}\)
\(\left( {\dfrac{{30}}{{35}} - \dfrac{{21}}{{35}}} \right):\dfrac{{2x}}{3} = - \dfrac{{11}}{{18}}\)
\(\dfrac{9}{{35}}:\dfrac{{2x}}{3} = - \dfrac{{11}}{{18}}\)
\(\dfrac{{2x}}{3} = \dfrac{9}{{35}}:\left( { - \dfrac{{11}}{{18}}} \right)\)
\(\dfrac{{2x}}{3} = \dfrac{9}{{35}}.\dfrac{{ - 18}}{{11}}\)
\(\dfrac{{2x}}{3} = \dfrac{{9.( - 18)}}{{35.11}}\)
\(\dfrac{2}{3}.x = \dfrac{{ - 162}}{{385}}\)
\(x = \dfrac{{ - 162}}{{385}}:\dfrac{2}{3}\)
\(x = \dfrac{{ - 162}}{{385}}.\dfrac{3}{2}\)
\(x = \dfrac{{ - 162.3}}{{385.2}}\)
\(x = \dfrac{{ - 81.2.3}}{{385.2}}\)
\(x = \dfrac{{ - 243}}{{385}}\)
Vậy \({x_0} = \dfrac{{ - 243}}{{385}} < \dfrac{{385}}{{385}} = 1\).
