## SUBJECT 2

Find the cardinality of the set:

$$
A=\left\{\overline{a b c} / \frac{a \sqrt{2}}{\sqrt{1}+\sqrt{2}+\sqrt{3}}-\frac{b \sqrt{3}}{\sqrt{1}+\sqrt{3}+\sqrt{4}}-\frac{2 c \sqrt{8}}{\sqrt{4}+\sqrt{8}+\sqrt{12}}=\frac{-3}{\sqrt{2}} ; a \neq b \neq c\right\}
$$

Prof. Nicolae Jurubiță
## SUBJECT 2

Rationalizing we get $a(1+\sqrt{2}-\sqrt{3})-b(\sqrt{3}-1)-c(2+2 \sqrt{2}-2 \sqrt{3})=-3 \sqrt{2} \ldots \ldots \ldots \ldots .3 p$

![](https://cdn.mathpix.com/cropped/2024_06_07_2aeee1df30a437c1f6afg-2.jpg?height=60&width=1465&top_left_y=1238&top_left_x=158)

![](https://cdn.mathpix.com/cropped/2024_06_07_2aeee1df30a437c1f6afg-2.jpg?height=57&width=1465&top_left_y=1288&top_left_x=158)

$a+b-2 c=0$ and $a-2 c=-3 \Rightarrow b=3$

$2 c=a+3$, a is odd, $a \in\{1,3,5,7\}$

$1 p$

Studying the cases and finding $A=\{132,534,735,936\}$ and $\operatorname{card} A=4$. $\qquad$