`\text{~~Phương~~}`
\(\dfrac{(x-5)}{(x+7)(x-2)}>0, ĐKXĐ: x\ne-7, x\ne2\\\to\left[ \begin{array}{l}\begin{cases}x-5>0\\(x+7)(x-2)>0\end{cases}\\\begin{cases}x-5<0\\(x+7)(x-2)<0\end{cases}\end{array} \right.\\\to \left[ \begin{array}{l}\begin{cases}x>5\\x\in(-∞,-7)∪(2,+∞)\end{cases}\\\begin{cases}x<5\\x\in(-7,2)\end{cases}\end{array} \right.\\\to\left[ \begin{array}{l}\in(5,+∞)\\x\in(-7,2)\end{array} \right. \\\to x\in(-7,2)∪(5,+∞)\)
