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Question Number 69201 by ahmadshah last updated on 21/Sep/19

	Answered by Kunal12588 last updated on 21/Sep/19
	$f(x)=((x−5)/(x^2 −25))=((x−5)/((x−5)(x+5))) ⇒(x−5)(x+5)≠0 ⇒x≠5,−5 ∴x∈R−{−5,5}$
	$f (x) = \frac{x - 5}{x^{2} - 25} = \frac{x - 5}{(x - 5) (x + 5)}$ $\Rightarrow (x - 5) (x + 5) \neq 0$ $\Rightarrow x \neq 5, - 5$ $∴ x \in R - {- 5, 5}$
	Commented by peter frank last updated on 21/Sep/19

	$r a n g e ? ?$
	Answered by Rio Michael last updated on 21/Sep/19
	$x^2 −25 = 0 (x −5)(x+5)=0 x= 5 or x =−5 x ∈R −{−5,5}$
	$x^{2} - 25 = 0$ $(x - 5) (x + 5) = 0$ $x = 5 o r x = - 5$ $x \in R - {- 5, 5}$
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