Question

Tìm ĐKXĐ của \(\sqrt{\dfrac{x^2}{2},}\sqrt{\dfrac{x-1}{-2}},\sqrt{x^2-4,}\sqrt{\dfrac{x-1}{2x^2}}\)

Answer 1

* \(\sqrt{\dfrac{x^2}{2}}\) . ĐKXĐ: mọi x

* \(\sqrt{\dfrac{x-1}{-2}}\) . ĐKXĐ: \(\dfrac{x-1}{-2}\ge0\Leftrightarrow x-1\le0\) (vì -2<0) <=> x \(\le\) 1

* \(\sqrt{x^2-4}\) . ĐKXĐ: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

* \(\sqrt{\dfrac{x-1}{2x^2}}\) .ĐKXĐ: \(\dfrac{x-1}{2x^2}\ge0\Leftrightarrow x-1\ge0\)(vì 2x^2 > 0 với mọi x) <=> x \(\ge\) 1