Câu 2:
a) Ta có: \(P=\dfrac{x-6}{x+3\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{x-6}{\sqrt{x}\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-6-\sqrt{x}-3+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-9}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
b) Ta có: \(x=\left(6\sqrt{0.5}-\sqrt{6}\right)\left(3\sqrt{2}+3\sqrt{6}-\sqrt{24}\right)\)
\(=\left(\sqrt{36\cdot0.5}-\sqrt{6}\right)\left(3\sqrt{2}+3\sqrt{6}-2\sqrt{6}\right)\)
\(=\left(\sqrt{18}-\sqrt{6}\right)\left(3\sqrt{2}+\sqrt{6}\right)\)
\(=\left(3\sqrt{2}-\sqrt{6}\right)\left(3\sqrt{2}+\sqrt{6}\right)\)
\(=18-6=12\)
Thay x=12 vào biểu thức \(P=\dfrac{\sqrt{x}-3}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{12}-3}{\sqrt{12}}=\dfrac{2\sqrt{3}-3}{2\sqrt{3}}=\dfrac{\sqrt{3}\left(2-\sqrt{3}\right)}{2\sqrt{3}}=\dfrac{2-\sqrt{3}}{2}\)
Vậy: Khi \(x=\left(6\sqrt{0.5}-\sqrt{6}\right)\left(3\sqrt{2}+3\sqrt{6}-\sqrt{24}\right)\) thì \(P=\dfrac{2-\sqrt{3}}{2}\)
