Question

Rezolvă orice si îți dau puncte​

Answer 1

Răspuns:

Rezolvare ex 1 de la subiectul I :

Arătați că √12 *(√3-3√2)+√8(3√3-√2) = 2

√12 *(√3-3√2)+√8(3√3-√2)

2√3*(√3-3√2)+2√2*(3√3-√2)

2√3*√3-2√3*3√2 + 2√3*3√3-2√2*√2

6-6√6+6√6-4

6-4

2

√12 *(√3-3√2)+√8(3√3-√2) = 2 Adevarat

Answer 2

$\it I)\\ \\ 2.\ \ f(x)=x^2-2x+a\\ \\ f(a)=2 \Rightarrow a^2-2a+a-2=0 \Rightarrow a(a-2)+(a-2)=0 \Rightarrow \\ \\ \Rightarrow (a-2)(a+1)=0 \Rightarrow a_1=-1,\ \ a_2=2\\ \\ \\ 3.\ \ \sqrt{9-x}=x-3 \Rightarrow \begin{cases}\ \it 9-x\geq0 \Rightarrow x\leq9\\ \\ \it x-3\geq0 \Rightarrow x\geq3\end{cases} \Rightarrow x\in[3,\ 9]\ \ \ \ \ \ (*)$

$\it \sqrt{9-x}=x-3 \Rightarrow (\sqrt{9-x})^2=(x-3)^2 \Rightarrow 9-x=x^2-6x+9\\ \\ \Rightarrow x^2-5x=0 \Rightarrow x(x-5)=0 \Rightarrow \begin{cases} \it x_1=0\ \stackrel{(*)}{\Longrightarrow}\ nu\ \ convine\\ \\ \it x-5=0 \Rightarrow x_2=5\ (solu\c{\it t}ia\ ecua\c{\it t}iei\ date)\end{cases}$