b, A=\(\dfrac{x}{x-4}\)+\(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

      =\(\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

      =\(\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

      =\(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

      =\(\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

      =\(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)}\)

x²-4x-5=0

⇔x²-5x+x-5=0

⇔(x²+x)-(5x+5)=0

⇔x(x+1)-5(x+1)=0

⇔(x-5)(x+1)=0

⇔\(\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

vậy phương trình có 2 nghiệm phân biệt x=5;x=-1