[tex]\sqrt{1+3x}=x-1\\ \\ \text{Conditii de existenta:}\\ \\ \begin{cases} 1+3x \geq 0\\ x-1 \geq 0\end{cases} \Rightarrow \begin{cases} 3x \geq -1\\ x\geq 1 \end{cases} \Rightarrow x \geq 1\Rightarrow D = [1,+\infty)\\ \\\\ \sqrt{1+3x}=x-1\Big|^2 \Rightarrow 1+3x = x^2-2x+1 \Rightarrow \\ \\\Rightarrow x^2-5x = 0 \Rightarrow x(x-5) = 0\\ \\\Rightarrow x = 0\geq 1\,\,(F)\quad \text{ sau }\quad x = 5\,\,(A) \Rightarrow \boxed{S = \{5\}}[/tex]
Restul se rezolvă la fel.
