Answer:
1.B
2.A
3. B
Step-by-step explanation:
1. 
We have the denominator of the fraction as following:

As the initial one is a fraction, so that its denominator has to be different from 0.
=> (
) ≠ 0
⇔ (x +1) (x +5) ≠ 0
⇔ (x + 1) ≠ 0; (x +5) ≠ 0
⇔ x ≠ -1; x ≠ -5
Replace it into the initial equation, we have:

As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)
=> 
So that
with x ≠ 1; x ≠ -5
So that the answer is B.
2. 
As the initial one is a fraction, so that its denominator has to be different from 0
=> x + 4 ≠ 0
=> x ≠ -4
As
is also a fraction, so that its denominator (x-1) has to be different from 0
=> x - 1 ≠ 0
=> x ≠ 1
We have an equation: 
=> 
Replace it into the initial equation, we have:

As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)
=> 
So that the initial equation is equal to
with x ≠-4; x ≠1
=> So that the correct answer is A
3. 
As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0
We have:
(4x + x^2) = 4x + x.x = x ( x + 4)
So that: (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0
⇔
⇔ 
As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:

As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:

So that
with x ≠ 0; x ≠ -4
=> The correct answer is B