Given as

f(x) = 2x + 5 and g(x) = x² + x

Here, both f(x) and g(x) are defined for all x ∈ R.

Therefore, domain of f = domain of g = R

(i) f + g

As we know, (f + g)(x) = f(x) + g(x)

(f + g)(x) = 2x + 5 + x² + x

= x² + 3x + 5

Now, (f + g)(x) is defined for all real numbers x.

∴ The domain of (f + g) is R

(ii) f – g

As we know, (f – g)(x) = f(x) – g(x)

(f – g)(x) = 2x + 5 – (x² + x)

= 2x + 5 – x² – x

= 5 + x – x²

(f – g)(x) is defined for all real numbers x.

∴ The domain of (f – g) is R

(iii) fg

As we know, (fg)(x) = f(x)g(x)

(fg)(x) = (2x + 5)(x² + x)

= 2x(x² + x) + 5(x² + x)

= 2x³ + 2x² + 5x² + 5x

= 2x³ + 7x² + 5x

(fg)(x) is defined for all real numbers x.

∴ The domain of fg is R

(iv) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = (2x+5)/(x²+x)
Here, (f/g) (x) is defined for all real values of x, except for the case when x² + x = 0.

x² + x = 0

x(x + 1) = 0

x = 0 or x + 1 = 0

x = 0 or –1

When x = 0 or –1, (f/g) (x) will be undefined as the division result will be indeterminate.

Thus, the domain of f/g = R – {–1, 0}