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Mình nghĩ đề câu a) là \(\frac{1}{1-\sqrt{x^2-3}}\) khi đó
\(1-\sqrt{x^2-3}\ne0\Rightarrow\sqrt{x^2-3}\ne1\Rightarrow x\ne\pm2\)và \(x^2-3\ge0\Leftrightarrow-\sqrt{3}\le x\le\sqrt{3}\)
b)
\(\sqrt{16-x^2}\ge0;\sqrt{2x+1}\ge0;\sqrt{x^2-8x+14}\ge0\)và \(\sqrt{2x+1}\ne0\)
\(\Leftrightarrow-4\le x\le4;x\ge-\frac{1}{2};4-\sqrt{2}\le x\le4+\sqrt{2};x\ne\frac{1}{2}\)
Như vậy \(-\frac{1}{2}< x\le4+\sqrt{2}\)
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/ \(\left\{{}\begin{matrix}-4\le x\le4\\x>-\frac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow-\frac{1}{2}< x\le4-\sqrt{2}\)
a, dk \(1-16x^2\ge0\Leftrightarrow\left(1-4x\right)\left(1+4x\right)\ge0\)
\(\Leftrightarrow-\frac{1}{4}\le x\le\frac{1}{4}\)
b tuong tu
c, \(\sqrt{\left(x-3\right)\left(5-x\right)}\ge0\Leftrightarrow\left(x-3\right)\left(5-x\right)\ge0\Leftrightarrow3\le x\le5\)
d.\(\sqrt{x^2-x+1}>0\)
ma \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
suy ra thoa man vs moi x
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/ \(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\\x^2-8x+8\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le4\\x>-\frac{1}{2}\\\left[{}\begin{matrix}x\ge4+2\sqrt{2}\\x\le4-2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-\frac{1}{2}< x\le4-2\sqrt{2}\)
\(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\\x^2-8x+14\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< x\le4-\sqrt{2}\)
xác định \(< =>\left\{{}\begin{matrix}\sqrt{16-x^2}\ge0\\\sqrt{2x+1}>0\\\sqrt{x^2-8x+14}\ge0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\le4-\sqrt{2}\\x\ge4_{ }+\sqrt{2}\end{matrix}\right.\\\end{matrix}\right.\)\(< =>-\dfrac{1}{2}< x\le4-\sqrt{2}\)
