Lời giải:
a)
ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ x-3\sqrt{x}-4\neq 0\\ \sqrt{x}+1\neq 0\\ \sqrt{x}-4\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (\sqrt{x}+1)(\sqrt{x}-4)\neq 0\\ \sqrt{x}+1\neq 0\\ \sqrt{x}-4\neq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ \sqrt{x}+1\neq 0\\ \sqrt{x}-4\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 16\end{matrix}\right.\)
b)
\(A=\frac{2(x+4)}{(\sqrt{x}+1)(\sqrt{x}-4)}+\frac{\sqrt{x}(\sqrt{x}-4)}{(\sqrt{x}+1)(\sqrt{x}-4)}-\frac{8(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-4)}\)
\(=\frac{2(x+4)+(x-4\sqrt{x})-(8\sqrt{x}+8)}{(\sqrt{x}+1)(\sqrt{x}-4)}=\frac{3x-12\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-4)}=\frac{3\sqrt{x}(\sqrt{x}-4)}{(\sqrt{x}+1)(\sqrt{x}-4)}=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
