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Question-69201




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}
$${f}\left({x}\right)=\frac{{x}−\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{25}}=\frac{{x}−\mathrm{5}}{\left({x}−\mathrm{5}\right)\left({x}+\mathrm{5}\right)} \\ $$$$\Rightarrow\left({x}−\mathrm{5}\right)\left({x}+\mathrm{5}\right)\neq\mathrm{0} \\ $$$$\Rightarrow{x}\neq\mathrm{5},−\mathrm{5} \\ $$$$\therefore{x}\in\mathbb{R}−\left\{−\mathrm{5},\mathrm{5}\right\} \\ $$
Commented by peter frank last updated on 21/Sep/19
range??
$${range}?? \\ $$
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}
$${x}^{\mathrm{2}} −\mathrm{25}\:=\:\mathrm{0} \\ $$$$\left({x}\:−\mathrm{5}\right)\left({x}+\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=\:\mathrm{5}\:{or}\:{x}\:=−\mathrm{5} \\ $$$${x}\:\in\mathbb{R}\:−\left\{−\mathrm{5},\mathrm{5}\right\} \\ $$

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