1: \(B=-\frac{1}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}+\frac{x+4}{x-9}\)
\(=\frac{-1}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}+\frac{x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-\sqrt{x}-3+\sqrt{x}-3+x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x-2}{x-9}\)
2: Để B nguyên thì x-2⋮x-9
=>x-9+7⋮x-9
=>7⋮x-9
=>x-9∈{1;-1;7;-7}
=>x∈{10;8;16;2}

