1.
[tex]\it a)\ 4-x\sqrt2<0 \Rightarrow 4<x\sqrt2 \Rightarrow x\sqrt2>4 \Rightarrow x\sqrt2>2\cdot2 \Rightarrow \\ \\ \Rightarrow x\sqrt2>2\cdor\sqrt2\cdot\sqrt2|_{:\sqrt2} \Rightarrow x>2\sqrt2 \Rightarrow x\in(2\sqrt2,\ \ \infty)\\ \\ \\ b)\ 2x^3-8x=2x(x^2-4)=2x(x^2-2^2)=2x(x-2)(x+2)\\ \\ c)\ \dfrac{x-2}{3}=\dfrac{4}{x+2} \Rightarrow (x-2)(x+2)=3\cdot4 \Rightarrow x^2-4=12|_{+4} \Rightarrow \\ \\ \\ \Rightarrow x^2=16\Rightarrow \sqrt{x^2}=\sqrt{16} \Rightarrow |x|=4 \Rightarrow x=\pm4;\ x=-4\ negativ\breve a[/tex]
2.
[tex]\it a)\ A=\{x\in\marhbb{R\Big|\ |x|<4}\}\\ \\ |x|<4 \Rightarrow -4<x<4 \Rightarrow A=(-4,\ \ 4)\\ \\ \\ b)\ A\cap\mathbb{Z}^*=\{-3,\ \ -2,\ \ -1,\ \ 1,\ \ 2,\ \ 3\}\\ \\ \\ c)\ p=\dfrac{nr.\ cazuri\ favorabile}{nr.\ cazuri\ posibile}=\dfrac{\ 2^{(2}}{6}=\dfrac{1}{3}[/tex]
3.
[tex]\it a)\ (x+2-\sqrt2)^2=\[[(x+2)-\sqrt2]^2=(x+2)^2-2\sqrt2(x+2)+(\sqrt2)^2=\\ \\ =x^2+4x+4-2\sqrt2x-4\sqrt2+2=x^2+(4-2\sqrt2)x+6-4\sqrt2\\ \\ b)\ \[[4x+20x^2:(-4x)]:(-x)=(4x-5x):(-x)=(-x):(-x)=1,\ \ \forall\ x \ne0\\ \\ c)\ -\sqrt4=-2\in\mathbb{Q};\ \sqrt{0,(1)}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\in\mathbb{Q};\ \sqrt{50}=\sqrt{25\cdot2}=5\sqrt2\in\mathbb{R\setminus\mathbb{Q}}\\ \\ Deci,\ \{-\sqrt4;\ \ \sqrt{0,(1)};\ \ \sqrt{50}\}\cap\mathbb{R}\setminus\mathbb{Q}=\{\sqrt{50}\}[/tex]
4.
[tex]\it a)\ BC||AD \Rightarrow (\widehat{VB,\ AD})=(\widehat{VB,\ BC})=60^o,\ deoarece\ \Delta VBC-echilateral\\ \\ b)\ BD=6\sqrt2\ cm\ (diagonala\ p\breve atratului \ ABCD),\ \ VB=VD=6\ cm\\ \\ Reciproca\ teoremei\ lui\ Pitagora \Rightarrow \Delta VBD-dreptunghic,\ m(\widehat V)=90^o.\\ \\ \mathcal{A}_{VBD}=\dfrac{c_1\cdot c_2}{2}=\dfrac{VB\cdot VD}{2}=\dfrac{6\cdot6}{2}=18\ cm^2\\ \\ \\ c)\ AD||BC;\ \ BC\subset (VBC)\ \Rightarrow AD||(VBC)[/tex]
