Answer:
[tex]f(g(x)) =\dfrac{1}{\sqrt{x - 2} - 5}[/tex]
[tex]D(f(g(x)) = (2;27) \cup (27; +\infty)[/tex]
Step-by-step explanation:
[tex]f(x) = \dfrac{1}{x - 5}[/tex]
[tex]g(x) = \sqrt{x - 2}[/tex]
[tex]f(g(x)) = \dfrac{1}{g(x) - 5} = \dfrac{1}{\sqrt{x - 2} - 5}[/tex]
[tex]D(f(g(x)):[/tex]
[tex]\displaystyle \left \{ {{x - 2 > 0} \atop { \sqrt{x - 2} - 5 \neq 0 }} \right.[/tex] [tex]\displaystyle \left \{ {{x > 2} \atop { \sqrt{x - 2} \neq 5 }} \right[/tex] [tex]\displaystyle \left \{ {{x > 2} \atop { (\sqrt{x - 2})^{2} \neq 5^{2} }} \right[/tex] [tex]\displaystyle \left \{ {{x > 2} \atop { x - 2 \neq 25 }} \right[/tex] [tex]\displaystyle \left \{ {{x > 2} \atop { x \neq 27 }} \right[/tex]
[tex]D(f(g(x)) = (2;27) \cup (27; +\infty)[/tex]
