Question

Tìm ĐKXĐ:

\(\dfrac{\sqrt{x-1}}{x^2}\), \(\sqrt{\dfrac{x}{\left(x-1\right)^2}}\) , \(\sqrt{x+5}-\sqrt{2x+1}\), \(\sqrt{3-x^2}\)

Answer 1

Giúp mk vs nha. Mk c.ơn

Answer 2

\(\dfrac{\sqrt{x-1}}{x^2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ge0\\x^2\ne0\end{matrix}\right.\Leftrightarrow x\ge1\)

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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-1\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ge0\)

\(\sqrt{x+5}-\sqrt{2x+1}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x+5\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Leftrightarrow x\ge\dfrac{-1}{2}\)

\(\sqrt{3-x^2}\)

ĐKXĐ: \(3-x^2\ge0\Leftrightarrow x\le\pm\sqrt{3}\)

Answer 3

\(\dfrac{\sqrt{x-1}}{x^2}\)

đkxđ :\(\left\{{}\begin{matrix}x-1\ge0\\x^2\ne0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

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

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\\left(x-1\right)^2\ne0\end{matrix}\right.\)