Question

Va rog din suflet!!!!

Answer 1

$\sqrt{x^2-6x+9} + 3|(x-3)(2x+1)| = 0 \\ \\ \sqrt{(x-3)^2} + 3|(x-3)(2x+1)| = 0 \\ \\ \underset{\geq 0}{\underbrace{|x-3|}} + \underset{\geq 0 }{\underbrace{3|(x-3)(2x+1)|}} = 0 \\ \\ \\ x-3 = 0 \quad \text{SI} \quad 3(x-3)(2x+1) = 0 \\ \\ x = 3 \quad \text{SI} \quad \Big(x=3 \quad \text{SAU} \quad 2x + 1 = 0\Big)\\ x = 3 \quad \text{SI} \quad \Big(x=3 \quad \text{SAU} \quad x = -\dfrac{1}{2}\Big) \\ \\ S = \{3\} \cap \left( \{3\} \cup \Big\{-\dfrac{1}{2}\Big\} \right)$

$S = \{3\} \cap \Big\{3,-\dfrac{1}{2}\Big\}\\ \\ \Rightarrow \boxed{S = \{3\}}$


(SI implica intersectie, iar SAU implica reuniune.)