[tex]\it a)\ E(x)=(x^2+x+2)(x^2+x-4)+8\\ \\ Vom\ nota\ x^2+x=t,\ iar\ expresia\ devine:\\ \\ E(t)=(t+2)(t-4)+8=t^2-4t+2t-8+8=t^2-2t=t(t-2)[/tex]
Revenim la notație:
[tex]\it E(x)=(x^2+x)(x^2+x-2)=x(x+1)(x^2+2x-x-2)=\\ \\ =x(x+1)[x(x+2)-(x+2)]=x(x+1)(x+2)(x-1) \Rightarrow \\ \\ \Rightarrow E(x)=(x-1)x(x+1)(x+2)[/tex]
[tex]\it F(x)=(x^2-x-2)(x^2-x-4)-8\\ \\ Vom\ nota\ x^2-x=t,\ iar\ expresia\ devine:\\ \\ F(t)=(t-2)(t-4)-8=t^2-4t-2t+8-8=t^2-6t=t(t-6)[/tex]
Revenim la notație:
[tex]\it F(x)=(x^2-x)(x^2-x-6)=x(x-1)(x^2-3x+2x-6)=\\ \\ =x(x-1)[x(x-3)+2(x-3)]=x(x-1)(x-3)(x+2)[/tex]
[tex]\it b)\ \dfrac{E(x)}{F(x)}= \dfrac{(x-1)x(x+1)(x+2)}{x(x-1)(x-3)(x+2)} =\dfrac{x+1}{x-3},\ \forall x\in\mathbb{R}\backslash \{-2,\ 0,\ 1\}[/tex]
[tex]\it c)\ x\in\mathbb{Z},\ \ \dfrac{E(x)}{F(x)}\in\mathbb{Z} \Rightarrow \dfrac{x+1}{x-3}\in\mathbb{Z} \Rightarrow x-3\ |\ x+1 \ \ \ \ \ (1)\\ \\ \\ Dar,\ \ x-3\ |\ x-3\ \ \ \ \ (2)\\ \\ \\ (1),\ (2)\Rightarrow x-3\ |\ x+1-x+3 \Rightarrow x-3\ |\ 4 \Rightarrow x-3\in D_4\Rightarrow[/tex]
[tex]\it \Rightarrow x-3\in\{\pm1,\ \pm2,\ \pm4\} \Rightarrow x-3\in\{-4,\ -2,\ -1,\ 1,\ 2,\ 4\}|_{+3}\Rightarrow\\ \\ \Rightarrow x\in\{-1,\ 1,\ 2,\ 4,\ 5,\ 7\}\\ \\ \\ Prin\ urmare\ \ A= \{-1,\ 1,\ 2,\ 4,\ 5,\ 7\}[/tex]
