Question

Subiectul al doilea fara 4 si 1

Answer 1

$\it 2)\ 1\in(m,\ n) \Leftrightarrow m<1<n\\ \\ m<1 \Rightarrow 3-2\sqrt2<1 \Rightarrow3-1<2\sqrt2 \Rightarrow 2<2\sqrt2\ (A)\ \ \ \ (1)\\ \\ 1<n \Rightarrow1<3+2\sqrt2 \Rightarrow 1-3<2\sqrt2 \Rightarrow-2<2\sqrt2\  (A)\ \ \ (2)\\ \\ (1),\ (2) \Rightarrow m<1<n \Leftrightarrow 1\in (m,\ n)$

$\it 3)\ 15-6\sqrt6 =9+6-6\sqrt6 =3^2-2\cdot3\sqrt6+(\sqrt6)^2=(3-\sqrt6)^2$

Ecuația devine:

$\it x\sqrt{(3-\sqrt6)^2}=2\sqrt6-6 \Rightarrow x(3-\sqrt6)=2\sqrt6-6 \Rightarrow\\ \\ \Rightarrow x(3-\sqrt6)= -6+2\sqrt6 \Rightarrow x(3-\sqrt6)=-2(3-\sqrt6)|_{:(3-\sqrt6)} \Rightarrow\\ \\ \Rightarrow x=-2=\dfrac{-2}{1}\in\mathbb{Q}$