Cộng, trừ, nhân, chia số hữu tỉ

$\dfrac{2}{3} + \dfrac{4}{5} = \dfrac{{10}}{{15}} + \dfrac{{12}}{{15}} = \dfrac{{22}}{{15}}.$

Với $x = \dfrac{a}{b};\,y = \dfrac{c}{d}\,\left( {b,d \ne 0} \right)$ ta có: $x.y = \dfrac{a}{b}.\dfrac{c}{d} = \dfrac{{a.c}}{{b.d}}$ .

Ta có $\dfrac{{ - 2}}{{13}} + \dfrac{{ - 11}}{{26}} = \dfrac{{ - 4}}{{26}} + \dfrac{{ - 11}}{{26}} = \dfrac{{ - 15}}{{26}}$

Do đó kết quả là số hữu tỉ âm.

Ta có $- \dfrac{6}{7}.\dfrac{{21}}{{12}} =  - \dfrac{6}{7}.\dfrac{7}{4} = \dfrac{{ - 6}}{4} =  - \dfrac{3}{2}$

Ta có:

$\dfrac{2}{3} + \dfrac{5}{4} = \dfrac{8}{{12}} + \dfrac{{15}}{{12}} = \dfrac{{23}}{{12}}.$

$\dfrac{1}{6} + \dfrac{3}{2} = \dfrac{1}{6} + \dfrac{9}{6} = \dfrac{{10}}{6} = \dfrac{5}{3}.$

$\dfrac{5}{3} + \dfrac{3}{2} = \dfrac{{10}}{6} + \dfrac{9}{6} = \dfrac{{19}}{6}.$

$1 + \dfrac{{13}}{{12}} = \dfrac{{12}}{{12}} + \dfrac{{13}}{{12}} = \dfrac{{25}}{{12}}.$

Do đó $\dfrac{{23}}{{12}}$ là kết quả của phép tính: $\dfrac{2}{3} + \dfrac{5}{4}.$

Ta có $\dfrac{5}{{11}}:\dfrac{{15}}{{22}}$$= \dfrac{5}{{11}}.\dfrac{{22}}{{15}} = \dfrac{{5.22}}{{11.15}} = \dfrac{2}{3}$

$\dfrac{{ - 3}}{{14}} = \dfrac{{7 - 10}}{{14}} = \dfrac{7}{{14}} - \dfrac{{10}}{{14}}$$= \dfrac{1}{2}-\dfrac{5}{7}$ nên C đúng

+) Đáp án B: $\dfrac{1}{{14}} - \dfrac{1}{7} = \dfrac{1}{{14}} - \dfrac{2}{{14}} = \dfrac{{ - 1}}{{14}}\ne \dfrac{{ - 3}}{{14}}$ nên loại B.

+) Đáp án A: $\dfrac{2}{3} - \dfrac{5}{7} = \dfrac{{14}}{{21}} - \dfrac{{15}}{{21}} = \dfrac{{ - 1}}{{21}}\ne \dfrac{{ - 3}}{{14}}$ nên loại A.

+) Đáp án D: $\dfrac{3}{{14}} - \dfrac{5}{{14}} = \dfrac{{ - 2}}{{14}} = \dfrac{{ - 1}}{7}\ne \dfrac{{ - 3}}{{14}}$ nên loại D.

Ta có $\dfrac{3}{2}.\dfrac{4}{7} = \dfrac{{3.4}}{{2.7}} = \dfrac{6}{7} > 0$

$\dfrac{2}{7} + \left( {\dfrac{{ - 3}}{5}} \right) + \dfrac{3}{5} = \dfrac{2}{7} + \left[ {\left( {\dfrac{{ - 3}}{5}} \right) + \dfrac{3}{5}} \right]$$= \dfrac{2}{7} + 0\, = \dfrac{2}{7}.$

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}$

$x + \dfrac{1}{2} = \dfrac{3}{4}$

$x\,\, = \dfrac{3}{4} - \dfrac{1}{2}$

$x = \dfrac{3}{4} - \dfrac{2}{4}$

$x = \dfrac{1}{4}$

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 $A < B$ .

Ta có $\dfrac{2}{5} + \left( { - \dfrac{4}{3}} \right) + \left( { - \dfrac{1}{2}} \right)$$= \dfrac{{12}}{{30}} + \left( {\dfrac{{ - 40}}{{30}}} \right) + \left( {\dfrac{{ - 15}}{{30}}} \right) = \dfrac{{12 - 40 - 15}}{{30}} = \dfrac{{ - 43}}{{30}}$

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 $x =  - \dfrac{3}{{16}}.$

Ta có $A = \dfrac{1}{3} - \left[ {\left( { - \dfrac{5}{4}} \right) - \left( {\dfrac{1}{4} + \dfrac{3}{8}} \right)} \right]$

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

$= \dfrac{1}{3} - \left[ {\left( { - \dfrac{5}{4}} \right) - \dfrac{5}{8}} \right]$

$= \dfrac{1}{3} - \left[ {\left( { - \dfrac{{10}}{8}} \right) - \dfrac{5}{8}} \right]$

$= \dfrac{1}{3} - \left( { - \dfrac{{15}}{8}} \right)$

$= \dfrac{1}{3} + \dfrac{{15}}{8}$

$= \dfrac{8}{{24}} + \dfrac{{45}}{{24}}$

$= \dfrac{{53}}{{24}}$

Vậy  $A = \dfrac{{53}}{{24}} > \dfrac{{48}}{{24}} = 2$  hay $A > 2$ .

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 $x =  - 1$ .

$\dfrac{2}{{11}} - \dfrac{5}{{13}} + \dfrac{9}{{11}} - \dfrac{8}{{13}} = \left( {\dfrac{2}{{11}} + \dfrac{9}{{11}}} \right) - \left( {\dfrac{5}{{13}} + \dfrac{8}{{13}}} \right) = \dfrac{{11}}{{11}} - \dfrac{{13}}{{13}} = 1 - 1 = 0.$

Ta có $\dfrac{5}{7}:x - \dfrac{2}{5} = \dfrac{1}{3}$

$\dfrac{5}{7}:x = \dfrac{1}{3} + \dfrac{2}{5}$

$\dfrac{5}{7}:x = \dfrac{5}{{15}} + \dfrac{6}{{15}}$

$\dfrac{5}{7}:x = \dfrac{{11}}{{15}}$

$x = \dfrac{5}{7}:\dfrac{{11}}{{15}}$

$x = \dfrac{5}{7}.\dfrac{{15}}{{11}}$

$x = \dfrac{{75}}{{77}}$

Vậy ${x_0} = \dfrac{{75}}{{77}} < \dfrac{{77}}{{77}} = 1$ .

$x + y = \dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad}}{{bd}} + \dfrac{{cb}}{{bd}} = \dfrac{{ad + cb}}{{bd}}.$

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