a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{1-\sqrt{5}+1}{\sqrt{5}-1+1}=\dfrac{2-\sqrt{5}}{\sqrt{5}}=\dfrac{2\sqrt{5}-5}{5}\)
b: Ta có: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
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