a) Thay x=25 vào A ta có:
\(A=\dfrac{2}{\sqrt{25}-1}=\dfrac{2}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}\)
b) \(B=\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}+\dfrac{x+12}{x-4}\) (Đk: \(x>0;x\ne4\))
\(B=\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}+\dfrac{x+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{x+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}-2-4\sqrt{x}-8+x+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c) Ta có: \(A\cdot B\)
\(=\dfrac{2}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
Ta có với \(x>0\)
Thì \(\sqrt{x}+2>2\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+2}< 2\)
\(\Rightarrow A\cdot B< 2\)
