Question

Exercițiul 2
DAU COROANA!!!

Answer 1

[tex]|2 \sqrt{3} -4|+|x \sqrt{3}|= \sqrt{25-9} \\ \\ 2 \sqrt{3}= \sqrt{3*2^{2}}= \sqrt{12} \ \textless \ \sqrt{16}=4 \rightarrow |2 \sqrt{3}-4|= 4-2 \sqrt{3} \\ \\ 4-2 \sqrt{3} +x \sqrt{3} = \sqrt{16} \rightarrow 4-2 \sqrt{3}+x \sqrt{3}=4 \rightarrow -2 \sqrt{3}+x \sqrt{3}=0 |: \sqrt{3} \\ \\ -2+x= 0 \rightarrow x=2 
|x \sqrt(3)| = 2 \sqrt{3} \rightarrow x \ apartine \ -2; 2 [/tex]

Answer 2

Analizăm primul modul :


[tex]\it \sqrt3\ \textless \ \sqrt4 \Rightarrow \sqrt3 \ \textless \ 2 \Rightarrow \sqrt3 - 2 \ \textless \ 0 |_{\cdot2}\Rightarrow 2\sqrt3-4\ \textless \ 0 \Rightarrow \\\;\\ \Rightarrow |2\sqrt3-4| = -2\sqrt3+4[/tex]

Analizăm membrul drept al ecuației:


$\it \sqrt{25-9} = \sqrt{16}=4$


Acum, ecuația se scrie :


[tex]\it -2\sqrt3 +4 +|x\sqrt3| = 4 \Rightarrow |x\sqrt3| = 4 -4 +2\sqrt3 \Rightarrow \\\;\\ \Rightarrow |x|\sqrt3 = 2\sqrt3|_{:\sqrt3} \Rightarrow |x |=2 \Rightarrow x = \pm2 \Rightarrow \begin{cases}\it x_1 = -2 \\\;\\ \it x_2 =\ \ 2\end{cases}[/tex]