Question

Tìm x để biểu thức có nghĩa:

a) \(\frac{1}{\sqrt{x-\sqrt{2x-1}}}\)

b) \(\frac{\sqrt{16-x^2}}{\sqrt{2x+1}}+\sqrt{x^2-8x+8}\)

Answer 1

ĐKXĐ:

a/ \(\left\{{}\begin{matrix}x-\sqrt{2x-1}>0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2x-1}-1\right)^2>0\\x\ge\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-1}\ne1\\x\ge\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge\frac{1}{2}\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\\x^2-8x+8\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le4\\x>-\frac{1}{2}\\\left[{}\begin{matrix}x\ge4+2\sqrt{2}\\x\le4-2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-\frac{1}{2}< x\le4-2\sqrt{2}\)