$\begin{array}{l}4)\\\text{- Ta có : $\dfrac8{12}=\dfrac23$}\\\text{- Do $\dfrac ab=\dfrac23$}\\\to\begin{cases} a=2m\\b=3m\end{cases}\quad(m\in\mathbb{Z};m\neq0)\\\text{mà $-4\leq a<17$}\\\to-4\leq2m<17\\\to -2\leq m<\dfrac{17}2\\\to -2\leq m<9\\\to m\in\{-2;-1;1;2;3;...8\}\\ \to \left[ \begin{array}{l}\begin{cases} a=2.(-2)=-4\\b=3.(-2)=-6 \end{cases}\\\begin{cases} a=2.(-1)=-2\\b=3.(-1)=-3 \end{cases}\\\dots\\\begin{cases} a=2.8=16\\b=3.8=24 \end{cases} \end{array}\right.\\\to\dfrac ab\in\left\{\dfrac{-4}{-6};\dfrac{-2}{-3};\dots;\dfrac{16}{24}\right\}\\\,\\5)\\a)\ \dfrac{x-1}7=\dfrac{-3}4\\\Leftrightarrow 4(x-1)=-3.7\\\Leftrightarrow 4x-4=-21\\\Leftrightarrow 4x=-17\\\Leftrightarrow x=\dfrac{-17}4\\\,\\b)\ \dfrac{-x}{20}=\dfrac{5}{-x}\\\Leftrightarrow (-x)^2=5.20\\\Leftrightarrow x^2=100\\\Leftrightarrow x^2=(\pm10)^2\\\Leftrightarrow x=\pm10\\\,\\c)\ \dfrac x3=\dfrac2y\\\Leftrightarrow xy=3.2\\\Leftrightarrow xy=6\\\text{mà $x,y\in\mathbb{Z}$}\\\to x,y\in Ư(6)=\{\pm1;\pm2;\pm3;\pm6\}\\\text{- Do $x>y$ và $xy=6$ nên ta có các cặp số $(x,y)$ thỏa mãn là :}\\(-2;-3);(-1;-6);(3;2);(6;1)\\\,\\d)\ \\\text{- Ta có : $x-y=5\to x=5+y$}\\\dfrac{x-4}{y-3}=\dfrac43\\\to 3(x-4)=4(y-3)\\\Leftrightarrow 3x-12=4y-12\\\Leftrightarrow 3x=4y\\\Leftrightarrow 3(5+y)=4y\\\Leftrightarrow 15+3y=4y\\\Leftrightarrow 3y-4y=-15\\\Leftrightarrow -y=-15\\\Leftrightarrow y=15\\\Leftrightarrow x=5+y=5+15=20\\\text{- Vậy $(x,y)=(20;25)$} \end{array}$
