a) \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
- Khi \(x=7+4\sqrt{3}\left(tm\right)\) thì:
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2+\sqrt{3}-1}{2+\sqrt{3}-2}=\dfrac{1+\sqrt{3}}{\sqrt{3}}=\dfrac{3+\sqrt{3}}{3}\)
b) \(B=\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\)
\(=\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(P=B:A=\dfrac{2-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
