How do you simplify #(x+5 )/ (x^2-25)#?

Answer 1

#1/(x-5)#

The key realization is that our denominator fits the difference of squares pattern #a^2-b^2#, which factors as #(a+b)(a-b)#.
We can rewrite #x^2-25# as #(x+5)(x-5)#, which allows us to rewrite our original expression as
#(x+5)/((x+5)(x-5))#
The #x+5# terms on the top and bottom cancel, and we're left with
#cancel(x+5)/(cancel(x+5)(x-5))#
#1/(x-5)#

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Answer 2

#1/(x-5)#

#x^2-25" is a "color(blue)"difference of squares"#
#a^2-b^2=(a-b)(a+b)#
#x^2-25=x^2-5^2=(x-5)(x+5)#
#(cancel((x+5)))/(cancel((x+5))(x-5))=1/(x-5)#
#"with restriction "x!=5#
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Answer 3

To simplify (x+5)/(x^2-25), we can factor the denominator as (x+5)(x-5). Then, we can cancel out the common factor of (x+5) in the numerator and denominator. The simplified form is 1/(x-5).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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