Lời giải:
a)
ĐKXĐ: \(\left\{\begin{matrix} 9-x^2\geq 0\\ x^2-1\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (3-x)(3+x)\geq 0\\ (x-1)(x+1)\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3\geq x\geq -3\\ \left[\begin{matrix} x\geq 1\\ x\leq -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} 3\geq x\geq 1\\ -3\leq x\leq -1\end{matrix}\right.\)
Vậy TXĐ \(D= [1;3]\cup [-3;-1]\)
b)
\(\left\{\begin{matrix} \frac{x^2-5x+4}{-2x^2+3x-1}\geq 0\\ -2x^2+3x-1\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{(x-4)(x-1)}{(x-1)(1-2x)}\geq 0\\ (x-1)(1-2x)\neq 0 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{x-4}{1-2x}\geq 0\\ x\neq 1; x\neq \frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 4\geq x> \frac{1}{2}\\ x\neq 1; x\neq \frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 4\geq x> \frac{1}{2}\\ x\neq 1\end{matrix}\right.\)
Vậy \(D=(\frac{1}{2}; 4]\setminus \left\{1\right\}\)
