Answer:
Third option and sixth option
Step-by-step explanation:
It is important to remember that, by definition, "Excluded values" are all those values that make the denominator equal to 0.
You need to substitute -5 into each rational expression:
[tex]\frac{x+5}{x-5}=\frac{x+5}{(-5)-5}=\frac{x+5}{-10}[/tex]
[tex]\frac{x^2-5}{x^2+5}=\frac{x^2-5}{(-5)^2+5}=\frac{x^2-5}{30}[/tex]
[tex]\frac{x-3}{x^2-25}=\frac{x-3}{(-5)^2-25}=\frac{x-3}{0}[/tex]
[tex]\frac{x^2-25}{2x^2+5}=\frac{x^2-25}{2(-5)^2+5}=\frac{x^2-25}{55}[/tex]
[tex]\frac{2x+1}{x^2+25}=\frac{2x+1}{(-5)^2+25}=\frac{2x+1}{50}[/tex]
[tex]\frac{(x-2)(x-5)}{(x+3)(x+5)}=\frac{(x-2)(x-5)}{(x+3)((-5)+5)}=\frac{(x-2)(x-5)}{0}[/tex]
