a) \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\) (ĐK: \(x>0;x\ne1\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+3}-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+3-3+\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{2\sqrt{x}}\)
\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
b) Ta có: \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{1}{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}=\dfrac{1}{\left(\sqrt{5}-1\right)^2}=\left(\dfrac{1}{\sqrt{5}-1}\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}+3}{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}}=3\sqrt{5}-2\)
c) Ta có: \(\dfrac{\sqrt{x}+3}{\sqrt{x}}=1+\dfrac{3}{\sqrt{x}}\)
\(\Rightarrow\sqrt{x}\in\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{1;9\right\}\)
