Nhân, chia số hữu tỉ

Ta có

$\begin{gathered}   - \frac{{12}}{{13}}x - 5 = 6\frac{1}{{13}}x \hfill \\   \Leftrightarrow  - \frac{{12}}{{13}}x - 5 = \left( {6 + \frac{1}{{13}}} \right)x \hfill \\   \Leftrightarrow  - 5 = \left( {6 + \frac{1}{{13}} + \frac{{12}}{{13}}} \right)x \hfill \\   \Leftrightarrow x =  - \frac{5}{7}. \hfill \\ \end{gathered}$

$\left( x+5 \right)\left( x-1 \right)=x.x-x.1+5.x-5={{x}^{2}}+4x-5.$

Ta có $\dfrac{3}{8}.19\dfrac{1}{3}-\dfrac{3}{8}.33\dfrac{1}{3}=-\dfrac{21}{4}$

Ta thấy: $\left( -\dfrac{1}{2} \right).\dfrac{5}{9}.\left( -\dfrac{7}{13} \right).\left( -\dfrac{3}{5} \right)\left( x\in \mathbb{Q} \right)$ luôn có kết quả là âm.

Vậy nếu $P > 0$ thì $x < 0.$

$P=0$ thì $x=0.$

$P < 0$ thì $x > 0.$

$6\dfrac{9}{11}:\left( -3 \right)=\left( 6+\dfrac{9}{11} \right):\left( -3 \right)=6:\left( -3 \right)+\dfrac{9}{11}:\left( -3 \right)=-2+\dfrac{-3}{11}=-2\dfrac{3}{11}.$

$\left( -\dfrac{1}{3}+\dfrac{5}{6} \right).11-7=\dfrac{1}{2}.11-7=-\dfrac{3}{2}.$

$\dfrac{7}{23}.\left[ \left( \dfrac{-8}{6}-\dfrac{45}{18} \right) \right]=\dfrac{7}{23}.\left( \dfrac{-8}{6}-\dfrac{15}{6} \right)=\dfrac{7}{23}.\dfrac{-23}{6}=-1\dfrac{1}{6}.$

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

$\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\Rightarrow \dfrac{3}{4}x=\dfrac{13}{14}\Rightarrow x=\dfrac{26}{21}=1\dfrac{5}{21}.$

$\dfrac{x}{y}=\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}:\dfrac{c}{d}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}$

Vậy $\dfrac{x}{y}$ là một số nguyên khi ad là bội của bc.

$1\dfrac{2}{5}.\dfrac{-3}{4}=\dfrac{7}{5}.\dfrac{-3}{4}=\dfrac{-21}{20};\,\,$

$1\dfrac{1}{17}.1\dfrac{1}{24}=\dfrac{18}{17}.\dfrac{25}{24}=\dfrac{75}{68};$

$4\dfrac{1}{5}:\left( -2\dfrac{4}{5} \right)=\dfrac{21}{5}.\dfrac{-5}{14}=\dfrac{-3}{2};\,$

$1\dfrac{4}{5}:\left( -\dfrac{3}{4} \right)=\dfrac{9}{5}.\left( \dfrac{-4}{3} \right)=\dfrac{-12}{5};$

Ta có $\left( -\dfrac{3}{7}+\dfrac{3}{5} \right):\dfrac{20}{21}+\left( -\dfrac{4}{7}+\dfrac{2}{5} \right):\dfrac{20}{21}=0$

Ta có $\left( \dfrac{2}{3}x-1 \right)\left( \dfrac{3}{4}x+\dfrac{1}{2} \right)=0\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{3}{2} \\ x=-\dfrac{2}{3} \end{array} \right.$

Ta có

$\begin{array}{l}\left( {{x^2} - \dfrac{1}{4}} \right)\left( {{x^2} + \dfrac{1}{9}} \right) = 0\\ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{{x^2} - \dfrac{1}{4} = 0}\\{{x^2} + \dfrac{1}{9} = 0\left( {VL} \right)}\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = \dfrac{{ - 1}}{2}}\\{x = \dfrac{1}{2}}\end{array}} \right.\end{array}$

Vậy có 2 giá trị của x thỏa mãn phương trình.

Ta có $\dfrac{2}{3}+\dfrac{5}{3}x=\dfrac{5}{7}\Leftrightarrow x=\dfrac{1}{35}$

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

Đặt $\dfrac{1}{x}=k\left( k\in \mathbb{Z} \right).$

Ta có: $x.\dfrac{1}{x}=x.k=1\Rightarrow x.k=1\Rightarrow x=\dfrac{1}{k}\left( k\in \mathbb{Z};k\ne 0 \right).$

Ta có $\dfrac{11}{5}.\dfrac{9}{12}:\left[ \left( -\dfrac{12}{15}.(-9) \right) \right]=\dfrac{11}{48}$

$x.x=x\Rightarrow x.x-x=0\Rightarrow x\left( x-1 \right)=0\Rightarrow x=0;\,\,x=1.$

Tổng các giá trị $x$ tìm được là: $0+1=1.$

Với $x=6,99;y=-1,01$ thì $A=12.\left[ 6,99-\left( -1,01 \right) \right]=12.\left( 6,99+1,01 \right)=12.8=96.$

Ta có $3\dfrac{1}{3}:2\dfrac{1}{2}-1 < x < 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\Leftrightarrow \dfrac{1}{3} < x < \dfrac{81}{14}$

Ta có $-x:\dfrac{3}{8}=\dfrac{8}{3}\Leftrightarrow x=-1$

Ta có $\dfrac{25}{2}.\left( -\dfrac{5}{7} \right)+\dfrac{3}{2}.\left( -\dfrac{5}{7} \right)=-10$

Ta có $\dfrac{-8}{11}.x=\dfrac{2}{5}.\dfrac{1}{4}\Leftrightarrow x=-\dfrac{11}{80}$

$\left( \dfrac{11}{12}:\dfrac{33}{16} \right).\dfrac{3}{5}=\dfrac{4}{15}$