\(B=\dfrac{x}{x-4}-\dfrac{1}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\left(x\ge0,x\ne4\right)\)
\(=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(C=\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\left(x>0,x\ne4\right)\)
\(=\dfrac{x-3\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(D=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{5}{1-\sqrt{x}}+\dfrac{4}{x-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
\(E=\dfrac{1}{2\sqrt{x}-3}-\dfrac{\sqrt{x}}{2\sqrt{x}+3}-\dfrac{3+8\sqrt{x}}{4x-9}\left(x\ge0,x\ne\dfrac{9}{4}\right)\)
\(=\dfrac{1}{2\sqrt{x}-3}-\dfrac{\sqrt{x}}{2\sqrt{x}+3}-\dfrac{3+8\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}+3-\sqrt{x}\left(2\sqrt{x}-3\right)-3-8\sqrt{x}}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}=\dfrac{-2x-3\sqrt{x}}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}\)
\(=\dfrac{-\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}=\dfrac{-\sqrt{x}}{2\sqrt{x}-3}\)
B = \(\dfrac{x}{x-4}-\dfrac{1}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
=\(\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
=1+\(\dfrac{2}{\sqrt{x}-2}\)