a: Thay x=25 vào A, ta được:
\(A=\frac{2}{\sqrt{25}-1}=\frac{2}{5-1}=\frac24=\frac12\)
b: \(B=\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}+\frac{x+12}{x-4}\)
\(=\frac{\sqrt{x}-2-4\left(\sqrt{x}+2\right)+x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+\sqrt{x}+10-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
c: \(A\cdot B-2\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}\cdot\frac{2}{\sqrt{x}-1}-2=\frac{2}{\sqrt{x}+2}-2\)
\(=\frac{2-2\sqrt{x}-4}{\sqrt{x}+2}=\frac{-2\sqrt{x}-2}{\sqrt{x}+2}<0\)
=>\(A\cdot B<2\)
