Answer:
1.B
2.A
3. B
Step-by-step explanation:
1. [tex]\frac{x+5}{x^{2} + 6x +5 }[/tex]
We have the denominator of the fraction as following:
[tex]x^{2} + 6x + 5 \\= x^{2} + (1 + 5)x + 5\\= x*x + 1x + 5x + 5*1\\= x ( x + 1) + 5(x + 1)\\= (x + 1) (x + 5)[/tex]
As the initial one is a fraction, so that its denominator has to be different from 0.
=> ([tex]x^{2} +6x+5[/tex]) ≠ 0
⇔ (x +1) (x +5) ≠ 0
⇔ (x + 1) ≠ 0; (x +5) ≠ 0
⇔ x ≠ -1; x ≠ -5
Replace it into the initial equation, we have:
[tex]\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)}[/tex]
As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)
=> [tex]\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}[/tex]
So that [tex]\frac{x+5}{x^{2} + 6x +5 } = \frac{1}{x+1}[/tex] with x ≠ 1; x ≠ -5
So that the answer is B.
2. [tex]\frac{(\frac{x^{2} -16 }{x-1} )}{x+4}[/tex]
As the initial one is a fraction, so that its denominator has to be different from 0
=> x + 4 ≠ 0
=> x ≠ -4
As [tex]\frac{x^{2}-16 }{x-1}[/tex] 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: [tex]x^{2} - y^{2} = (x - y ) (x+y)[/tex]
=> [tex]x^{2} - 16 = x^{2} - 4^{2} = (x -4) (x +4)[/tex]
Replace it into the initial equation, we have:
[tex]\frac{(\frac{x^{2} -16 }{x-1} )}{x+4} \\= \frac{x^{2} -16 }{x-1} . \frac{1}{x + 4}\\= \frac{(x-4)(x+4)}{x-1}. \frac{1}{x + 4}[/tex]
As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)
=> [tex]\frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}[/tex]
So that the initial equation is equal to [tex]\frac{x-4}{x-1}[/tex] with x ≠-4; x ≠1
=> So that the correct answer is A
3. [tex]\frac{x}{4x + x^{2} }[/tex]
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
⇔ [tex]\left \{ {{x\neq 0} \atop {(x+4)\neq0 }} \right.[/tex] ⇔ [tex]\left \{ {{x\neq 0} \atop {x \neq -4 }} \right.[/tex]
As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:
[tex]\frac{x}{4x + x^{2} } = \frac{x}{x(x+4)}[/tex]
As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:
[tex]\frac{x}{x(x+4)} =\frac{x/x}{x(x+4)/x} = \frac{1}{x+4}[/tex]
So that [tex]\frac{x}{4x+x^{2} } = \frac{1}{x+4}[/tex] with x ≠ 0; x ≠ -4
=> The correct answer is B
