Question

Ajutor pls exercițiul 20 b
Dau coroana

Answer 1

Explicație pas cu pas:

a) x ≠0, y ≠0, x ≠ y

$\sqrt{\overline {1,(x)} + \overline {4,(y)}} = \sqrt{1 + \dfrac{x}{9} + 4 + \dfrac{y}{9}} = \\ = \sqrt{5 + \dfrac{x + y}{9}} = \dfrac{ \sqrt{45 + x + y} }{ \sqrt{9} } \\ = \dfrac{ \sqrt{45 + x + y} }{3} \in \mathbb{Q}$

$\implies \sqrt{45 + x + y} \in \mathbb{N}$

$1 \leqslant x \leqslant 9 \\ 1 \leqslant y \leqslant 9 \\ 2 + 45 \leqslant x + y + 45\leqslant 18 + 45 \\ 47 \leqslant x + y + 45 \leqslant 63$

$\implies x + y + 45 = 49 \iff x + y = 4$

x = 1 => y = 3

x = 3 => y = 1

b)

$\sqrt{\overline {0,x(y)} + \overline {0,y(x)}} = \sqrt{ \dfrac{\overline {xy} - x}{90} + \dfrac{\overline {yx} - y}{90}} = \\ = \sqrt{ \dfrac{10x + y - x + 10y + x - y}{90}} = \sqrt{ \dfrac{10x + 10y}{90}} \\ = \sqrt{ \dfrac{10(x + y)}{90}} = \sqrt{ \dfrac{x + y}{9}} = \dfrac{ \sqrt{x + y} }{3} \in \mathbb{N}$

$\implies \sqrt{x + y} \in \mathcal{M}_{3} - \Big\{0\Big\} \iff \\ \sqrt{x + y} \in \Big\{ 3; 6; ... \Big\} \iff x + y \in \Big\{ 9; 36; ... \Big\}$

$1 \leqslant x \leqslant 9 \\ 1 \leqslant y \leqslant 9 \\ 2 \leqslant x + y \leqslant 18 \\ 2 < 9 < 18$

$\implies x + y = 9$

x = 1 => y = 8

x = 2 => y = 7

x = 3 => y = 6

...

x = 8 => y = 1