Question

tìm đkxđ của \(\sqrt{\dfrac{1}{x^4-16}}\)

Answer 1

ĐKXĐ: x^4-16>0

=>(x-2)(x+2)>0

=>x>2 hoặc x<-2

Answer 2

\(\sqrt{\dfrac{1}{x^4-16}}=\sqrt{\dfrac{1}{\left(x^2\right)^2-\left(2^2\right)^2}}\)

\(=\sqrt{\dfrac{1}{\left(x^2-2^2\right)\left(x^2+2^2\right)}}=\sqrt{\dfrac{1}{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}}\)

Vì \(x^2+4\ge4\forall x\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)

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