[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]
