Answer:
The correct answer options are C. [tex]\frac{x^2+5}{x-8}[/tex] and D. [tex]\frac{x^2-x-56}{x^2-64}[/tex].
Step-by-step explanation:
The values which make the denominator equal to zero are called the excluded values.
Here, we can substitute 8 for x and check if it makes the denominator 0.
[tex]\frac{x-8}{x+8} = \frac{8-8}{8+8} =\frac{0}{16} =0[/tex]
[tex]\frac{x-2}{x^2-4} = \frac{8-2}{8^2-4} =\frac{6}{60} =\frac{1}{10}[/tex]
[tex]\frac{x^2+5}{x-8} = \frac{8^2+5}{8-8} =\frac{69}{0}[/tex]
[tex]\frac{x^2-x-56}{x^2-64} = \frac{8^2-8-56}{8^2-64} = \frac{0}{0} =0[/tex]
[tex]\frac{8x^2-2}{x^2-16} = \frac{8(8)^2-2}{8^2-16} =\frac{510}{48}[/tex]
