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Va rog ajutati-ma la aceste exercitii,sunt obosita si maine trebuie sa am tema facuta va rog dau puncte multe

Va Rog Ajutatima La Aceste Exercitiisunt Obosita Si Maine Trebuie Sa Am Tema Facuta Va Rog Dau Puncte Multe class=

Răspuns :

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]