Question

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Answer 1

$\it 4.\\ \\ \underbrace{(\it3x^2+2)}_{>0}(|x-2|-1)\leq0 \Rightarrow |x-2|-1\leq0 \Rightarrow |x-2|\leq1 \Rightarrow \\ \\ \\ \Rightarrow -1\leq x-2\leq1|_{+2} \Rightarrow 1\leq x\leq3\stackrel{x\in\mathbb{Z}}{\Longrightarrow}\ x\in\{-1,\ 0,\ 1,\ 2,\ 3\}$

$\it 5.\\ \\ \\\dfrac{5-|2x+3|}{|x-1|}\geq0 \Rightarrow \begin{cases} \it 5-|2x+3|\geq0\ \ \ \ \ \ \ (1)\\ \\ x-1\ne0 \Rightarrow \it x\ne1\ \ \ \ \ (2)\end{cases}\\ \\ \\ (1) \Rightarrow |2x+3|\leq5 \Rightarrow -5\leq2x+3\leq5|_{-3} \Rightarrow -8\leq 2x\leq2|_{:2}\ \Rightarrow \\ \\ \Rightarrow -4\leq x\leq1\ \stackrel{(2)}{\Longrightarrow}\ x\in[-4,\ \ 1)$