[tex]\sqrt{x-1} = x-3 \\ \\ \boxed{1}\quad x-1\geq 0 \Rightarrow x
\geq 1 \\ \boxed{2} \quad x-3\geq 0 \Rightarrow x \geq 3 \\ $Din
\boxed{1} \cap $ $ \boxed{2} \Rightarrow D = [3,+\infty)[/tex]
[tex]\sqrt{x-1} = x-3\Big|^2 \Rightarrow \Big(\sqrt{x-1}\Big)^2 = (x-3)^2 \Rightarrow |x-1| = x^2-6x+9 \\ x \geq 1 \Rightarrow |x-1| = +(x-1) = x-1 \\ \\ \Rightarrow x-1 = x^2-6x+9 \Rightarrow x^2-6x-x+9+1 = 0 \Rightarrow \\ \Rightarrow x^2-7x+10 = 0 \\ \\ \Delta = (-7)^2-4\cdot 1\cdot 10 \Rightarrow \Delta = 49 - 40 \Rightarrow \Delta = 9\Rightarrow x_{1,2} = \dfrac{-(-7)\pm \sqrt{9} }{2\cdot 1} [/tex]
[tex]
\Rightarrow x_{1,2} = \dfrac{7\pm 3 }{2}\Rightarrow x_1 = 2\notin D,\quad x_2 = 5 \in D \\ \\ \Rightarrow \boxed{S = \big\{5\big\}}[/tex]
