1) Ta có: \(A=\left(\dfrac{3}{x-3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)+x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{3\sqrt{x}+9+x-3\sqrt{x}}{x}\)
\(=\dfrac{x+9}{x}\)
2) Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{6-2\sqrt{5}+9}{6-2\sqrt{5}}=\dfrac{14-2\sqrt{5}}{6-2\sqrt{5}}\)
\(=\dfrac{\left(14-2\sqrt{5}\right)\left(6+2\sqrt{5}\right)}{16}\)
\(=\dfrac{84+28\sqrt{5}-12\sqrt{5}-20}{16}\)
\(=\dfrac{64+16\sqrt{5}}{16}=4+\sqrt{5}\)
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