Question

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Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it a)\ 1\circ 1=1+1-6\cdot1\cdot1=1+1-6=-4\\ \\ b)\ m\circ(1-m)=m+1-m-6m(1-m)=1-6m+6m^2=6m^2-6m+1\\ \\ Inecua\c{\it t}ia\ se\ scrie:\\ \\ 6m^2-6m+1 < 3 \Big|_{-3}\Rightarrow 6m^2-6m-2 < 0\Big|_{:2} \Rightarrow 3m^2-3m-1 < 0$

$\it Fie \ func\c{\it t}ia\ \ real\breve a\ f(m)=3m^2-3m-1.\\ \\ f(m) < 0\ pentru\ m\in(m_1,\ \ m_2),\ \ unde\ m_1,\ m_2\ sunt\ r\breve ad\breve acinile\\ \\ ecua\c{\it t}iei\ f(m)=0\ .\\ \\ \\ f(m)=0 \Rightarrow 3m^3-3m-1=0 \\ \\ \Delta=9+12=21;\ \ m_{1,2}=\dfrac{3\pm\sqrt{21}}{6};\ \ f(m) < 0 \Rightarrow m\in\Big(\dfrac{3-\sqrt{21}}{6};\ \dfrac{3+\sqrt{21}}{6}\Big)$

$\it \sqrt{21} < \sqrt{25} \Rightarrow \sqrt{21} < 5\Big|_{\cdot(-1)} \Rightarrow -\sqrt{21} > -5\Big|_{+3} \Rightarrow 3-\sqrt{21} > -2\Big|_{:6} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{3-\sqrt{21}}{6} > -\dfrac{1}{3}\ \ \ \ \ \ (1)\\ \\ \\ \sqrt{21} < \sqrt{25} \Rightarrow \sqrt{21} < 5\Big|_{+3} \Rightarrow 3+\sqrt{21} < 8\Big|_{:6} \Rightarrow \dfrac{3+\sqrt{21}}{6} < \dfrac{4}{3}=1\dfrac{1}{3}\ \ \ (2)$

$\it (1),\ (2) \Rightarrow \Big(\dfrac{3-\sqrt{21}}{6},\ \ \dfrac{3+\sqrt{21}}{6}\Big)\subset\Big(-\dfrac{1}{3},\ \ 1\dfrac{1}{3}\Big)\\ \\ \\ m\in\Big(-\dfrac{1}{3},\ \ 1\dfrac{1}{3}\Big)\ \stackrel{m\in\mathbb{Z}}{\Longrightarrow}\ \ m\in\{0,\ 1\}$

$\it c)\ \ x\circ x=-4 \Rightarrow x+x-6x^2=-4 \Rightarrow 6x^2-2x-4=0\Big|_{:2} \Rightarrow 3x^2-x-2=0 \Rightarrow \\ \\ \\ \Rightarrow 3x^2-3x+2x-2=0 \Rightarrow 3x(x-1)+2(x-1)=0 \Rightarrow (x-1)(3x+2)=0 \Rightarrow \\ \\ \\ \Rightarrow \begin{cases}\ \it x-1=0 \Rightarrow x=1\\ \\ \it 3x+2=0 \Rightarrow x=-\dfrac{2}{3}\ \not\in\mathbb{N}\end{cases}\\ \\ \\ Ecua\c{\it t}ia\ admite\ \hat\imath n\ \mathbb{N}\ solu\c{\it t}ia\ x=1$