Question

oricare din 8,9 si 11

Answer 1

9)

$a= \sqrt{x+49}+\sqrt{x+625},\quad x\in [0,51]\\ \\ 0\leq x\leq 51 \Big|+49 \Rightarrow 49\leq x+49 \leq 100\Big|\sqrt{}\Rightarrow \\ \\\text{Putem scrie sub radical inegalitatea deoarece }\sqrt{x} \text{ e functie strict}\\ \text{crescatoare pe }[0,+\infty) \\\\ \Rightarrow \sqrt{49}\leq \sqrt{x+49}\leq \sqrt{100}\Rightarrow 7\leq\sqrt{x+49}\leq 10 \,\, (*)$

$0\leq x\leq 51\Big|+625 \Rightarrow 625 \leq x+625 \leq 676\Big|\sqrt{} \Rightarrow \\\\ \Rightarrow \sqrt{625} \leq\sqrt{x+625}\leq \sqrt{676} \Rightarrow 25\leq \sqrt{x+625}\leq 26\,\,(**) \\ \\ \text{Din }(*) + (**) \Rightarrow 7+25 \leq\sqrt{x+49}+\sqrt{x+625} \leq 10+26 \Rightarrow\\ \\\Rightarrow 32\leq\sqrt{x+49}+\sqrt{x+625}\leq 36\Rightarrow a\in [32,36]$

11)

$\sqrt{3-2\sqrt 2} = \sqrt{(1-\sqrt 2)^2} = |1-\sqrt 2| = \sqrt 2 - 1 = \\ \\ =-1+1\cdot \sqrt 2,\quad -1,1 \in \mathbb{Z} \Rightarrow -1+1\cdot \sqrt 2 \in A$