[tex]\it a)\ \ E(x) =(2x+5)^2-16\\ \\ (2x+5)^2\geq0|_{-16} \Rightarrow (2x+5)^2-16\geq -16 \Rightarrow E(x)\geq-16 \Rightarrow \\ \\ \Rightarrow minE(x)=-16,\ pentru\ oricare\ x\in\mathbb{R}[/tex]
[tex]\it b)\ \ x\in\mathbb{N} \Rightarrow x\geq0|_{\cdot2}\Rightarrow2x\geq0|_{+5}\Rightarrow 2x+5\geq5 \Rightarrow (2x+5)^2\geq25|_{-16}\Rightarrow\\ \\ \Rightarrow (2x+5)^2-16\geq 9 \Rightarrow E(x)\geq9 \Rightarrow minE(x)=9,\ pentru\ oricare\ x\in\mathbb{N}[/tex]
[tex]\it c)\ \ E(x)=(2x+5)^2-16 = (2x+5)^2-4^2=(2x+5-4)(2x+5+4)=\\ \\ =(2x+1)(2x+9)\ \ \ (*)\\ \\ x=0 \stackrel{(*)}{\Longrightarrow} E(0) =1\cdot9=9=\ num\breve{a}r\ compus \\ \\ x>0 \Rightarrow 2x+1>1,\ 2x+9>1\stackrel{(*)}{\Longrightarrow} E(x)=\ num\breve{a}r\ compus,\ x\in\mathbb{N}[/tex]
[tex]\it d)\ \ (*) \Rightarrow |E(m)|=|2m+1|\cdot|2m+9|\ \ \ (**)\\ \\ (**)\Rightarrow |E(m)|-\ num\breve{a}r\ prim\ dac\breve{a}\begin{cases} \it |2x+1|=1,\ |2x+9| = num\breve{a}r\ prim\\ \\ sau\\ \\ \it |2x+9|=1,\ |2x+1| = num\breve{a}r\ prim \end{cases}[/tex]
[tex]\it A)\ \ |2x+1|=1\Rightarrow 2x+1=\pm1 \Rightarrow 2x+1\in\{-1,\ 1\}|_{-1} \Rightarrow 2x\in\{-2,\ 0\}|_{:2}\Rightarrow \\ \\ \Rightarrow x\in\{-1,\ 0\}\\ \\ I)\ \ m=-1\Rightarrow |E(-1)|=|-2+1|\cdot|-2+9|=|-1|\cdot|7|=7\ (num\breve{a}r\ prim)\\ \\ II)\ \ m=0\Rightarrow |E(0)|=|1|\cdot|9|=9\ (num\breve{a}r\ compus)[/tex]
[tex]\it B)\ \ |2m+9|=1\Rightarrow 2m+9\in\{-1,\ 1\}|_{-9}\Rightarrow 2m\in\{-10,\ -8\}|_{:2}\Rightarrow \\ \\ \Rightarrow m\in\{-5,\ -4\} \\ \\ I)\ \ m=-5\Rightarrow |E(-5)|=|-10+1|\cdot|-10+9| =|-9|\cdot|-1|=9\cdot1=\\ \\ =9\ (num\breve{a}r\ compus)\\ \\ II)\ m=-4\Rightarrow |E(-4)|=|-8+1|\cdot|-8+9|=|-7|\cdot|1|=7\ (num\breve{a}r\ prim)[/tex]
[tex]\it Deci,\ pentru\ m\in\{-1,\ -4\} \Rightarrow |E(m)|=7\ (num\breve{a}r\ prim)[/tex]
