a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne16\end{cases}}\)
\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)
\(\Leftrightarrow B=\frac{2x+8+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
b) Để B nguyên'
\(\Leftrightarrow3\sqrt{x}⋮\sqrt{x}+1\)
\(\Leftrightarrow3\left(\sqrt{x}+1\right)-3⋮\sqrt{x}+1\)
\(\Leftrightarrow3⋮\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)(Đã loại những giá trị âm)
\(\Leftrightarrow x\in\left\{0;4\right\}\)
Vậy để \(B\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)
Loại giá trị 0 ở câu b cho mik nhé (Vì ktm đkxđ)