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UPenn2544
29.03.2020 •
Mathematics
Simplify the following:
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Ответ:
1.B
2.A
3. B
Step-by-step explanation:
1.![\frac{x+5}{x^{2} + 6x +5 }](/tpl/images/0569/3579/c24ca.png)
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)
=>![\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}](/tpl/images/0569/3579/fe02f.png)
So that
with x ≠ 1; x ≠ -5
So that the answer is B.
2.![\frac{(\frac{x^{2} -16 }{x-1} )}{x+4}](/tpl/images/0569/3579/2206a.png)
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:![x^{2} - y^{2} = (x - y ) (x+y)](/tpl/images/0569/3579/7cb7b.png)
=>![x^{2} - 16 = x^{2} - 4^{2} = (x -4) (x +4)](/tpl/images/0569/3579/fac43.png)
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)
=>![\frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}](/tpl/images/0569/3579/70a0b.png)
So that the initial equation is equal to
with x ≠-4; x ≠1
=> So that the correct answer is A
3.![\frac{x}{4x + x^{2} }](/tpl/images/0569/3579/5d35d.png)
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
⇔
⇔ ![\left \{ {{x\neq 0} \atop {x \neq -4 }} \right.](/tpl/images/0569/3579/2ffa9.png)
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
Ответ: