Question

tìm đk:

a,\(\sqrt{3x^2+1}\)

b,\(\sqrt{\dfrac{x+1}{x-2}}\)

Answer 1

\(a.\sqrt{3x^2+1}\) xác định \(< =>3x^2+1\ge0\) mà \(3x^2+1>0\)

nên \(\sqrt{3x^2+1}\) luôn xác định \(\forall x\in R\)

\(b,\sqrt{\dfrac{x+1}{x-2}}\) xác định\(< =>\dfrac{x+1}{x-2}\ge0\)

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\x-2< 0\end{matrix}\right.\end{matrix}\right.< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)

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