Nhân chia các số hữu tỉ

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 có một giá trị của $x$ thỏa mãn điều kiện.

Ta có: $A = \dfrac{3}{4}.\dfrac{5}{9} + \dfrac{6}{7}:\dfrac{4}{3} - 1\dfrac{2}{5}:1\dfrac{1}{3}$

$= \dfrac{3}{4}.\dfrac{5}{9} + \dfrac{6}{7}:\dfrac{4}{3} - \dfrac{7}{5}:\dfrac{4}{3}$

$= \dfrac{3}{4}.\dfrac{5}{9} + \dfrac{6}{7}.\dfrac{3}{4} - \dfrac{7}{5}.\dfrac{3}{4}$

$= \dfrac{3}{4}.\left( {\dfrac{5}{9} + \dfrac{6}{7} - \dfrac{7}{5}} \right)$

$= \dfrac{3}{4}.\left( {\dfrac{{175}}{{315}} + \dfrac{{270}}{{315}} - \dfrac{{441}}{{315}}} \right)$

$= \dfrac{3}{4}.\dfrac{{175 + 270 - 441}}{{315}}$

$= \dfrac{3}{4}.\dfrac{4}{{315}}$

$= \dfrac{{3.4}}{{4.315}}$

$= \dfrac{{3.4}}{{4.3.105}}$

$= \dfrac{1}{{105}}$

Vậy $A = \dfrac{1}{{105}}$.

+) $x:\left( { - 2\dfrac{1}{{15}}} \right) + 3\dfrac{1}{2} =  - \dfrac{3}{4}$

$x:\dfrac{{ - 31}}{{15}} + \dfrac{7}{2} =  - \dfrac{3}{4}$

$x:\dfrac{{ - 31}}{{15}} =  - \dfrac{3}{4} - \dfrac{7}{2}$

$x:\dfrac{{ - 31}}{{15}} = \dfrac{{ - 3}}{4} - \dfrac{{14}}{4}$

$x:\dfrac{{ - 31}}{{15}} = \dfrac{{ - 17}}{4}$

$x = \dfrac{{ - 17}}{4}.\dfrac{{ - 31}}{{15}}$

$x = \dfrac{{ - 17.( - 31)}}{{4.15}}$

$x = \dfrac{{527}}{{60}}$

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 ${x_2} = \dfrac{6}{{17}}$  

Ta có: ${x_1} = \dfrac{{527}}{{60}} > \dfrac{{60}}{{60}} = 1$; ${x_2} = \dfrac{6}{{17}} < \dfrac{{17}}{{17}} = 1$. Do đó ${x_2} < {x_1}$. 

Ta có: $\left( { - \dfrac{5}{8}} \right) - x:3\dfrac{5}{6} + 7\dfrac{3}{4} =  - 2$

$\left( { - \dfrac{5}{8}} \right) - x:\dfrac{{23}}{6} + \dfrac{{31}}{4} =  - 2$

 $x:\dfrac{{23}}{6} = \left( { - \dfrac{5}{8}} \right) + \dfrac{{31}}{4} + 2$

 $x:\dfrac{{23}}{6} = \dfrac{{ - 5}}{8} + \dfrac{{62}}{8} + \dfrac{{16}}{8}$

$x:\dfrac{{23}}{6} = \dfrac{{73}}{8}$

$x = \dfrac{{73}}{8}.\dfrac{{23}}{6}$

$x = \dfrac{{1679}}{{48}}$

$A = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{{15}}{{119}}}}{{\dfrac{{ - 10}}{{68}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{{15}}{{238}}}}$

$= \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{{3.5}}{{7.17}}}}{{\dfrac{{ - 2.5}}{{4.17}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{{3.5}}{{14.17}}}}$

$= \dfrac{{\dfrac{1}{2}.\dfrac{5}{{17}} - \dfrac{{13}}{{14}}.\dfrac{5}{{17}} + \dfrac{3}{7}.\dfrac{5}{{17}}}}{{\dfrac{{ - 1}}{2}.\dfrac{5}{{17}} + \dfrac{{26}}{{14}}.\dfrac{5}{{17}} - \dfrac{3}{{14}}.\dfrac{5}{{17}}}}$

$= \dfrac{{\dfrac{5}{{17}}.\left( {\dfrac{1}{2} - \dfrac{{13}}{{14}} + \dfrac{3}{7}} \right)}}{{\dfrac{5}{{17}}.\left( {\dfrac{{ - 1}}{2} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}} \right)}}$  

$= \dfrac{{\dfrac{1}{2} - \dfrac{{13}}{{14}} + \dfrac{3}{7}}}{{\dfrac{{ - 1}}{2} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}}}$

$= \dfrac{{\dfrac{7}{{14}} - \dfrac{{13}}{{14}} + \dfrac{6}{{14}}}}{{\dfrac{{ - 7}}{{14}} + \dfrac{{26}}{{14}} - \dfrac{3}{{14}}}}$

$= \dfrac{{\dfrac{0}{{14}}}}{{\dfrac{{16}}{{14}}}}$

$= 0$

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 $x = \dfrac{{ - 1}}{{15}}$; $x = \dfrac{{ - 14}}{3}$.

Tổng hai giá trị trên là: $\dfrac{{ - 1}}{{15}} + \dfrac{{ - 14}}{3} = \dfrac{{ - 1}}{{15}} + \dfrac{{ - 70}}{{15}} = \dfrac{{( - 1) + ( - 70)}}{{15}} = \dfrac{{ - 71}}{{15}}$.

Ta có: $\dfrac{2}{9}.\left[ {\left( { - \dfrac{5}{{11}}:\dfrac{{13}}{8} - \dfrac{5}{{11}}:\dfrac{{13}}{5}} \right) + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\left( {\dfrac{{ - 5}}{{11}}.\dfrac{8}{{13}} - \dfrac{5}{{11}}.\dfrac{5}{{13}}} \right) + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\left( {\dfrac{{ - 5.8}}{{11.13}} - \dfrac{{5.5}}{{11.13}}} \right) + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\left( {\dfrac{{ - 40}}{{143}} - \dfrac{{25}}{{143}}} \right) + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\dfrac{{ - 65}}{{143}} + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\dfrac{{ - 5}}{{11}} + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\left[ {\dfrac{{ - 15}}{{33}} + \dfrac{{ - 1}}{{33}}} \right] + \dfrac{{ - 3}}{4}$

$= \dfrac{2}{9}.\dfrac{{ - 16}}{{33}} + \dfrac{{ - 3}}{4}$

$= \dfrac{{2.( - 16)}}{{9.33}} + \dfrac{{ - 3}}{4}$

$= \dfrac{{ - 32}}{{297}} + \dfrac{{ - 3}}{4}$

$= \dfrac{{ - 128}}{{1188}} + \dfrac{{ - 891}}{{1188}}$

$= \dfrac{{ - 1019}}{{1188}}$

$- \dfrac{{12}}{5}.\dfrac{3}{4} = \dfrac{{\left( { - 12} \right).3}}{{5.4}} = \dfrac{{ - 36}}{{20}} = \dfrac{{ - 9}}{5}$.

$\dfrac{5}{{13}}:\dfrac{7}{{26}} = \dfrac{5}{{13}}.\dfrac{{26}}{7} = \dfrac{{5.26}}{{13.7}} = \dfrac{{5.13.2}}{{13.7}} = \dfrac{{10}}{7}$.

$\dfrac{{20}}{3}:\left({\dfrac{{-5}}{9}}\right)=\dfrac{{20}}{3}.\left({\dfrac{{-9}}{5}}\right)=\dfrac{{20.(-9)}}{{3.5}}=\dfrac{{4.5.3.(-3)}}{{3.5}}=-12$.

$\dfrac{{17}}{5}:\dfrac{{34}}{9}=\dfrac{{17}}{5}.\dfrac{9}{{34}}=\dfrac{{17.9}}{{5.34}}=\dfrac{{17.9}}{{5.2.17}}=\dfrac{9}{{10}}$.

$\dfrac{{ - 5}}{9}:2\dfrac{1}{2} = \dfrac{{ - 5}}{9}:\dfrac{5}{2} = \dfrac{{ - 5}}{9}.\dfrac{2}{5} = \dfrac{{ - 2}}{9}$.

Ta có $x:\left( {\dfrac{2}{5} - 1\dfrac{2}{5}} \right) = -2$

$x:\left( {\dfrac{2}{5} - \dfrac{7}{5}} \right) = -2$

$x:\left( {\dfrac{{ - 5}}{5}} \right) = -2$

$x:\left( { - 1} \right) = -2$

$x = (-1).\left( { - 2} \right)$

$x =   2$

Vậy $x =  2$ .

Ta có: $x:\left( {\dfrac{2}{9} - \dfrac{1}{5}} \right) = \dfrac{-13}{{26}}$

$x:\left( {\dfrac{{10}}{{45}} - \dfrac{9}{{45}}} \right) = \dfrac{-1}{2}$

$x:\dfrac{1}{{45}} = \dfrac{-1}{2}$

$x = \dfrac{-1}{2}.\dfrac{1}{{45}}$

$x = \dfrac{{-1.1}}{{2.45}}$

$x = \dfrac{-1}{{90}}$

Vậy $x = \dfrac{-1}{{90}}$.

Ta có:

$\dfrac{4}{5}.\dfrac{{-15}}{2}=\dfrac{{4.(-15)}}{{5.2}}=\dfrac{{-60}}{{10}}=-6$

Ta thấy$-6$ là số nguyên âm.

$\dfrac{{ - 15}}{{13}}.\dfrac{{13}}{7} = \dfrac{{( - 15).13}}{{13.7}} = \dfrac{{ - 15}}{7}$.

Ta thấy $\dfrac{{ - 15}}{7}$ là số nhỏ hơn $0$.

Với: $x = \dfrac{m}{n};\,y = \dfrac{p}{q}\,\left( {n,q \ne 0} \right)$ thì tích $x.y$ bằng:

$x.y = \dfrac{m}{n}.\dfrac{p}{q} = \dfrac{{m.p}}{{n.q}}$ .

Cách chia hai số hữu tỉ: Viết chúng dưới dạng phân số rồi áp dụng quy tắc chia hai phân số: chia phân số thứ nhất cho phân số thứ hai bằng cách nhân phân số thứ nhất với nghịch đảo của phân số thứ hai.

$\left( {\dfrac{{ - 2}}{3} + \dfrac{3}{7}} \right):\dfrac{4}{5} + \left( {\dfrac{{ - 1}}{3} + \dfrac{4}{7}} \right):\dfrac{4}{5}$

$= \left( {\dfrac{{ - 2}}{3} + \dfrac{3}{7}} \right).\dfrac{5}{4} + \left( {\dfrac{{ - 1}}{3} + \dfrac{4}{7}} \right).\dfrac{5}{4}$

$= \left( {\dfrac{{ - 2}}{3} + \dfrac{3}{7} + \dfrac{{ - 1}}{3} + \dfrac{4}{7}} \right).\dfrac{5}{4}$

$\begin{array}{l} = \left[ {\left( {\dfrac{{ - 2}}{3} + \dfrac{{ - 1}}{3}} \right) + \left( {\dfrac{3}{7} + \dfrac{4}{7}} \right)} \right].\dfrac{5}{4}\\ = \left( { - 1 + 1} \right).\dfrac{5}{4}\\ = 0\end{array}$