Bài 4:
a: \(P=\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}+2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right):\left(1-\dfrac{3\sqrt{x}-9}{x-9}\right)\)
\(=\left(\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right):\dfrac{x-9-3\sqrt{x}+9}{x-9}\)
\(=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{x-9}{x-3\sqrt{x}}\)
\(=\dfrac{-x+9+x-4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b: \(x=\dfrac{\sqrt{4+2\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}\)
\(=\dfrac{3-1}{1}=\dfrac{2}{1}=2\)
Thay x=2 vào P, ta được:
\(P=\dfrac{2+\sqrt{2}}{\sqrt{2}}=\sqrt{2}+1\)
