Question

Va rog repede, nu înțeleg nimic

Answer 1

[tex]E(x)=\frac{x^3-4x}{x^2-x-6}\cdot \left(\frac{x^2-4}{x^2+x-12} \right)^{-1}+\frac{4}{x+2}\\ a)\text{Mai intai punem conditiile de existenta(numitorul nu are voie sa}\\ \text{fie 0):}\\ x^2-x-6\neq 0\Rightarrow (x-3)(x+2)\neq 0\Rightarrow x\notin \{3,-2\}\\ x^2-4\neq 0\Rightarrow x^2\neq 4\Rightarrow x\notin \{\pm 2\}\\ x+2\neq 0\Rightarrow x\neq -2\\ Sa\ revenim:\\ E(x)=\frac{x(x^2-4)}{(x-3)(x+2)}\cdot \frac{(x+4)(x-3)}{x^2-4}+\frac{4}{x+2}\\ E(x)=\frac{x(x+4)}{x+2}+\frac{4}{x+2}\\ E(x)=\frac{x^2+4x+4}{x+2}[/tex]
[tex]E(x)=\frac{(x+2)^2}{x+2}\Rightarrow E(x)=x+2\forall x\in \mathbb{R} -\{\pm 2,3\}\\ b)1\ \textless \ |E(x)|\ \textless \ 3\\ 1\ \textless \ |x+2|\ \textless \ 3\\ \text{Luam fiecare caz in parte si rezolvam:}\\ i)1\ \textless \ |x+2|\\ x+2\ \textgreater \ 1\Rightarrow x\in (-1,\infty)\\ Sau\ x+2\ \textless \ -1\Rightarrow x\in(-\infty,-3)\\ S_1=(-\infty,-3)\cup(-1,\infty)\\ ii)|x+2|\ \textless \ 3\\ -3\ \textless \ x+2\ \textless \ 3|-2\\ -5\ \textless \ x\ \textless \ 1\\ S_2=(-5,1)\\ Intersectand\ S_1\ si\ S2\ avem\ ca:x\in (-5,-3)\cup (-1,1)\\ Deoarece\ x\in \mathbb{Z}\Rightarrow x\in \{-4,0\}\\ Solutia\ finala\ :S:x \in \{-4,0\}[/tex]