Question

Tìm ĐKXĐ

a. \(3-\sqrt{1-16x^2}\)

b. \(\frac{1}{1-\sqrt{x^2-3}}\)

c.\(\sqrt{8x-x^2-15}\)

d. \(\frac{2}{\sqrt{x^2-x+1}}\)

e. \(A=\frac{1}{\sqrt{x-\sqrt{2x-1}}}\)

g. \(\frac{\sqrt{16-x^2}}{\sqrt{2x+1}}+\sqrt{x^2-8x+14}\)

Answer 1

a/ \(1-16x^2\ge0\Rightarrow x^2\le16\Rightarrow-\frac{1}{4}\le x\le\frac{1}{4}\)

b/ \(\left\{{}\begin{matrix}x^2-3\ge0\\x^2-3\ne1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\\x\ne\pm2\end{matrix}\right.\)

c/ \(8x-x^2-15\ge0\Rightarrow3\le x\le5\)

d/ Hàm số xác định với mọi x

e/ \(\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x\ne1\end{matrix}\right.\)

f/ \(\left\{{}\begin{matrix}-4\le x\le4\\x>-\frac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow-\frac{1}{2}< x\le4-\sqrt{2}\)