Các dạng toán về phép nhân và phép chia phân số

Phép nhân phân số cũng có các tính chất tương tự phép nhân số tự nhiên như tính chất giao hoán, tính chất kết hợp, tính chất nhân phân phối.

Phân số nghịch đảo của số $- 3$ là $\dfrac{1}{{ - 3}}$

$\dfrac{1}{{12}} \cdot \dfrac{8}{{ - 9}} = \dfrac{{1.8}}{{12.\left( { - 9} \right)}}$$= \dfrac{{1.2.4}}{{4.3.\left( { - 9} \right)}} = \dfrac{2}{{ - 27}} = \dfrac{{ - 2}}{{27}}$

$\dfrac{{\left( { - 7} \right)}}{6}:\left( { - \dfrac{{14}}{3}} \right) = \dfrac{{ - 7}}{6}.\dfrac{{ - 3}}{{14}} = \dfrac{{1.1}}{{2.2}} = \dfrac{1}{4}$

Đáp án A: $\dfrac{2}{7}.\dfrac{{14}}{6} = \dfrac{{2.14}}{{7.6}} = \dfrac{{28}}{{42}} = \dfrac{2}{3}$ nên A đúng.

Đáp án B: $25.\dfrac{{ - 4}}{{15}} = \dfrac{{25.\left( { - 4} \right)}}{{15}} = \dfrac{{ - 100}}{{15}} = \dfrac{{ - 20}}{3}$ nên B đúng.

Đáp án C: ${\left( {\dfrac{2}{{ - 3}}} \right)^2}.\dfrac{9}{4} = \dfrac{{{2^2}}}{{{{\left( { - 3} \right)}^2}}}.\dfrac{9}{4}$$= \dfrac{4}{9}.\dfrac{9}{4} = 1$ nên C đúng.

Đáp án D: $\dfrac{{ - 16}}{{25}}.\left( {\dfrac{{25}}{{ - 24}}} \right) = \dfrac{{ - 16}}{{25}}.\dfrac{{25}}{{ - 24}}$$= \dfrac{{ - 2}}{{ - 3}} = \dfrac{2}{3} \ne  - \dfrac{2}{3}$ nên D sai.

$\begin{array}{l}\left( { - \dfrac{3}{5}} \right).x = \dfrac{4}{{15}}\\x = \dfrac{4}{{15}}:\left( {\dfrac{{ - 3}}{5}} \right)\\x = \dfrac{4}{{15}}.\dfrac{5}{{ - 3}}\\x =  - \dfrac{4}{9}\end{array}$

$\dfrac{{ - 5}}{6}.\dfrac{{120}}{{25}} < x < \dfrac{{ - 7}}{{15}}.\dfrac{9}{{14}}$

$\dfrac{{ - 5}}{6}.\dfrac{{24}}{5} < x < \dfrac{{ - 1}}{5}.\dfrac{3}{2}$

$- 4 < x < \dfrac{{ - 3}}{10}$

$x \in \left\{ { - 3; - 2; - 1} \right\}$

$\begin{array}{l}M = \dfrac{5}{6}:{\left( {\dfrac{5}{2}} \right)^2} + \dfrac{7}{{15}}\\M = \dfrac{5}{6}:\dfrac{{25}}{4} + \dfrac{7}{{15}}\\M = \dfrac{5}{6}.\dfrac{4}{{25}} + \dfrac{7}{{15}}\\M = \dfrac{{1.2}}{{3.5}} + \dfrac{7}{{15}}\\M = \dfrac{2}{{15}} + \dfrac{7}{{15}}\\M = \dfrac{9}{{15}} = \dfrac{3}{5}\end{array}$

Khi đó $a = 3,b = 5$ nên $a + b = 8$

Vì $x$ nguyên dương nên $x > 0$

mà ${\left( {\dfrac{{ - 5}}{3}} \right)^3} = \dfrac{{ - 125}}{{27}} < 0$ nên 

${\left( {\dfrac{{ - 5}}{3}} \right)^3} < 0 < x <\dfrac{{ - 24}}{{35}}.\dfrac{{ - 5}}{6}$

Khi đó:

$0 < x < \dfrac{{ - 24}}{{35}}.\dfrac{{ - 5}}{6}$

$0 < x < \dfrac{4}{7}$

Vì $\dfrac{4}{7} < 1$ nên $0 < x < 1$ nên không có số nguyên dương nào thỏa mãn.

$P = \left( {\dfrac{7}{{20}} + \dfrac{{11}}{{15}} - \dfrac{{15}}{{12}}} \right):\left( {\dfrac{{11}}{{20}} - \dfrac{{26}}{{45}}} \right)$

$P = \left( {\dfrac{{21}}{{60}} + \dfrac{{44}}{{60}} - \dfrac{{75}}{{60}}} \right):\left( {\dfrac{{99}}{{180}} - \dfrac{{104}}{{180}}} \right)$

$P = \dfrac{{ - 10}}{{60}}:\dfrac{{ - 5}}{{180}} = \dfrac{{ - 10}}{{60}}.\dfrac{{180}}{{ - 5}} = 6$

 $Q = \dfrac{{5 - \dfrac{5}{3} + \dfrac{5}{9} - \dfrac{5}{{27}}}}{{8 - \dfrac{8}{3} + \dfrac{8}{9} - \dfrac{8}{{27}}}}:\dfrac{{15 - \dfrac{{15}}{{11}} + \dfrac{{15}}{{121}}}}{{16 - \dfrac{{16}}{{11}} + \dfrac{{16}}{{121}}}}$

$Q = \dfrac{{5\left( {1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{{27}}} \right)}}{{8\left( {1 - \dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{{27}}} \right)}}:\dfrac{{15\left( {1 - \dfrac{1}{{11}} + \dfrac{1}{{121}}} \right)}}{{16\left( {1 - \dfrac{1}{{11}} + \dfrac{1}{{121}}} \right)}}$

$Q = \dfrac{5}{8}:\dfrac{{15}}{{16}} = \dfrac{5}{8}.\dfrac{{16}}{{15}} = \dfrac{2}{3}$

Vì $6 > \dfrac{2}{3}$ nên $P > Q$

$\begin{array}{l}x\;:\;\dfrac{5}{8} = \dfrac{{ - 14}}{{35}} \cdot \dfrac{{15}}{{ - 42}}\\x:\dfrac{5}{8} = \dfrac{{ - 2}}{5}.\dfrac{5}{{ - 14}}\\x:\dfrac{5}{8} = \dfrac{1}{7}\\x = \dfrac{1}{7}.\dfrac{5}{8}\\x = \dfrac{5}{{56}}\end{array}$

$\left( {x + \dfrac{1}{4} - \dfrac{1}{3}} \right):\left( {2 + \dfrac{1}{6} - \dfrac{1}{4}} \right) = \dfrac{7}{{46}}$

$\left( {x + \dfrac{1}{4} - \dfrac{1}{3}} \right):\dfrac{{23}}{{12}} = \dfrac{7}{{46}}$

$x + \dfrac{1}{4} - \dfrac{1}{3} = \dfrac{7}{{46}}.\dfrac{{23}}{{12}}$

$x + \dfrac{1}{4} - \dfrac{1}{3} = \dfrac{7}{{24}}$

$x = \dfrac{7}{{24}} - \dfrac{1}{4} + \dfrac{1}{3}$

$x = \dfrac{3}{8}$

$\left( {\dfrac{7}{6} + x} \right):\dfrac{{16}}{{25}} = \dfrac{{ - 5}}{4}$

$\dfrac{7}{6} + x = \dfrac{{ - 5}}{4}.\dfrac{{16}}{{25}}$

$\dfrac{7}{6} + x = \dfrac{{ - 1}}{1}.\dfrac{4}{5}$

$\dfrac{7}{6} + x = \dfrac{{ - 4}}{5}$

$x = \dfrac{{ - 4}}{5} - \dfrac{7}{6}$

$x = \dfrac{{ - 59}}{{30}}$

$\dfrac{{13}}{{15}} - \left( {\dfrac{{13}}{{21}} + x} \right).\dfrac{7}{{12}} = \dfrac{7}{{10}}$

$\left( {\dfrac{{13}}{{21}} + x} \right).\dfrac{7}{{12}} = \dfrac{{13}}{{15}} - \dfrac{7}{{10}}$

$\left( {\dfrac{{13}}{{21}} + x} \right).\dfrac{7}{{12}} = \dfrac{1}{6}$

$\dfrac{{13}}{{21}} + x = \dfrac{1}{6}:\dfrac{7}{{12}}$

$\dfrac{{13}}{{21}} + x = \dfrac{2}{7}$

$x = \dfrac{2}{7} - \dfrac{{13}}{{21}}$

$x =  - \dfrac{1}{3}$

$M = \dfrac{{17}}{5}.\dfrac{{ - 31}}{{125}}.\dfrac{1}{2}.\dfrac{{10}}{{17}}.{\left( {\dfrac{{ - 1}}{2}} \right)^3}$

$M = \dfrac{{17.\left( { - 31} \right).1.10.{{\left( { - 1} \right)}^3}}}{{{{5.125.2.17.2}^3}}}$

$M = \dfrac{{ - 31.\left( { - 1} \right)}}{{{{125.2}^3}}}$

$M = \dfrac{{31}}{{1000}}$

$N = \left( {\dfrac{{17}}{{28}} + \dfrac{{28}}{{29}} - \dfrac{{19}}{{30}} - \dfrac{{20}}{{31}}} \right).\left( {\dfrac{{ - 5}}{{12}} + \dfrac{1}{4} + \dfrac{1}{6}} \right)$

$N = \left( {\dfrac{{17}}{{28}} + \dfrac{{28}}{{29}} - \dfrac{{19}}{{30}} - \dfrac{{20}}{{31}}} \right).\left( {\dfrac{{ - 5}}{{12}} + \dfrac{3}{{12}} + \dfrac{2}{{12}}} \right)$

$N = \left( {\dfrac{{17}}{{28}} + \dfrac{{28}}{{29}} - \dfrac{{19}}{{30}} - \dfrac{{20}}{{31}}} \right).0$

$N = 0$

Vậy $M + N = \dfrac{{31}}{{1000}} + 0 = \dfrac{{31}}{{1000}}$

$\dfrac{{5x}}{3}:\dfrac{{10{x^2} + 5x}}{{21}}$ $= \dfrac{{5x}}{3}.\dfrac{{21}}{{10{x^2} + 5x}}$ $= \dfrac{{5x.21}}{{3.5x.\left( {2x + 1} \right)}}$ $= \dfrac{7}{{2x + 1}}$

Để biểu thức đã cho có giá trị là số nguyên thì $\dfrac{7}{{2x + 1}}$ nguyên

Do đó $2x + 1 \in Ư\left( 7 \right) = \left\{ { \pm 1; \pm 7} \right\}$

Ta có bảng:

Vậy $x \in \left\{ {0; - 1;3; - 4} \right\}$ suy ra có $4$ giá trị thỏa mãn.

$B = \dfrac{{{2^2}}}{3} \cdot \dfrac{{{3^2}}}{8} \cdot \dfrac{{{4^2}}}{{15}} \cdot \dfrac{{{5^2}}}{{24}} \cdot \dfrac{{{6^2}}}{{35}} \cdot \dfrac{{{7^2}}}{{48}} \cdot \dfrac{{{8^2}}}{{63}} \cdot \dfrac{{{9^2}}}{{80}}$

$= \dfrac{{2.2}}{{1.3}} \cdot \dfrac{{3.3}}{{2.4}} \cdot \dfrac{{4.4}}{{3.5}} \cdot \dfrac{{5.5}}{{4.6}} \cdot \dfrac{{6.6}}{{5.7}} \cdot \dfrac{{7.7}}{{6.8}} \cdot \dfrac{{8.8}}{{7.9}} \cdot \dfrac{{9.9}}{{8.10}}$

$= \dfrac{{2.3.4.5.6.7.8.9}}{{1.2.3.4.5.6.7.8.}} \cdot \dfrac{{2.3.4.5.6.7.8.9}}{{3.4.5.6.7.8.9.10}}$

$= \dfrac{9}{1} \cdot \dfrac{2}{{10}} = \dfrac{{9.2}}{{1.10}} = \dfrac{9}{5}$

Quãng đường AB là: $40.\dfrac{5}{4} = 50$ (km)

Thời gian người đó đi từ B về A là: $\dfrac{{50}}{{45}} = \dfrac{{10}}{9}$ (giờ)

Gọi phân số lớn nhất cần tìm là: $\dfrac{a}{b}$ ($a;b$ là nguyên tố cùng nhau)

Ta có: $\dfrac{{12}}{{35}}:\dfrac{a}{b} = \dfrac{{12b}}{{35{\rm{a}}}}$ là số nguyên, mà $12;35$ là nguyên tố cùng nhau

Nên $12 \vdots a;b \vdots 35$

Ta lại có: $\dfrac{{18}}{{49}}:\dfrac{a}{b} = \dfrac{{18b}}{{49{\rm{a}}}}$ là số nguyên, mà $18$ và $49$ nguyên tố cùng nhau

Nên $18 \vdots a;b \vdots 49$

Để $\dfrac{a}{b}$ lớn nhất ta có $a = UCLN(12;18) = 6$ và $b = BCNN(35;49) = 245$

Vậy tổng $a + b = 6 + 245 = 251$

$\begin{array}{l}\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{x\left( {x + 1} \right):2}} = \dfrac{{2019}}{{2021}}\\2.\left[ {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{x(x + 1)}}} \right] = \dfrac{{2019}}{{2021}}\\2.\left( {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right) = \dfrac{{2019}}{{2021}}\\2.\left( {\dfrac{1}{2} - \dfrac{1}{{x + 1}}} \right) = \dfrac{{2019}}{{2021}}\\1 - \dfrac{2}{{x + 1}} = \dfrac{{2019}}{{2021}}\\\dfrac{2}{{x + 1}} = 1 - \dfrac{{2019}}{{2021}}\\\dfrac{2}{{x + 1}} = \dfrac{2}{{2021}}\\x + 1 = 2021\\x = 2020\end{array}$