1) a)
\(y=\frac{\sqrt{4-x}+\sqrt{x+3}}{\left(\left|x\right|-1\right)\sqrt{x^2-2x+1}}\\ ĐK:\left[{}\begin{matrix}4-x\ge0\\x+3\ge0\\\left|x\right|-1\ne0\\x^2-2x+1>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\le4\\x\ge-3\\x\ne\pm1\\\left(x-1\right)^2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le4\\x\ge-3\\x\ne\pm1\\x\ne1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-3\le x\le4\\x\ne\pm1\end{matrix}\right.\\ TXĐ:D=\left[-3;4\right]\backslash\left\{-1;1\right\}\)
\(b.\\ y=\frac{\sqrt{x^2-6x+9}+\sqrt{\left|x\right|-2}}{\left(x^4-4x^2+3\right)\left(\sqrt{x}-2\right)}\\ ĐK:\left\{{}\begin{matrix}x^2-6x+9\ge0\\\left|x\right|-2\ge0\\x^4-4x^2+3\ne0\\\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-2\ne0\end{matrix}\right.\end{matrix}\right. \)
(tương tự câu a)
2)
\(y=f\left(x\right)=\frac{x^4-6x^2+2}{\left|x\right|-1}\\ ĐK:\left|x\right|-1\ne0\Leftrightarrow x\ne\pm1\\ TXĐ:D=R\backslash\left\{-1;1\right\}\\ \forall x\in D\Rightarrow-x\in D\)
Ta có: f(-x)=\(\frac{\left(-x\right)^4-6\left(-x\right)^2+2}{\left|-x\right|-1}=\frac{x^4-6x^2+2}{\left|x\right|-1}\)
=f(x)
⇒Hàm số đã cho là hàm số chẵn
