\(a,A=\dfrac{\sqrt{\dfrac{1}{4}}+2}{\sqrt{\dfrac{1}{4}}}=\dfrac{\dfrac{1}{2}+2}{\dfrac{1}{2}}=\dfrac{\dfrac{1+4}{2}}{\dfrac{1}{2}}=5\)
\(b,B=\dfrac{x+8\sqrt{x}}{x-4}+\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{-1-\sqrt{x}}{\sqrt{x}-2}\\ =\dfrac{x+8\sqrt{x}+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(1+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{x-4}\\ =\dfrac{x+8\sqrt{x}+x-\sqrt{x}-2\sqrt{x}+2-\left(\sqrt{x}+x+2+2\sqrt{x}\right)}{x-4}\\ =\dfrac{2x+5\sqrt{x}+2-3\sqrt{x}-x-2}{x-4}\\ =\dfrac{x+2\sqrt{x}}{\left(x-4\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
a: Khi x=1/4 thì \(A=\dfrac{5}{2}:\dfrac{1}{2}=5\)
b: \(B=\dfrac{x+8\sqrt{x}+x-3\sqrt{x}+2-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
