a) Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\), ta được:
\(A=\left(\sqrt{\dfrac{1}{4}}+1\right):\left(\sqrt{\dfrac{1}{4}}-2\right)\)
\(\Leftrightarrow A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)\)
\(\Leftrightarrow A=\dfrac{3}{2}:\dfrac{-3}{2}\)
\(\Leftrightarrow A=\dfrac{3}{2}\cdot\dfrac{2}{-3}=\dfrac{3}{-3}=-1\)
Vậy: Khi \(x=\dfrac{1}{4}\) thì A=-1
b) Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}-3\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)-3\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
Lời giải:a) Với \(x=\frac{1}{4}\Rightarrow \sqrt{x}=\frac{1}{2}\)
Khi đó: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=-1\)
b) \(B=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}-8}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{\sqrt{x}-8}{(\sqrt{x}-2)(\sqrt{x}-3)}\)
\(=\frac{x-4+\sqrt{x}-8}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{x+\sqrt{x}-12}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{(\sqrt{x}-3)(\sqrt{x}+4)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{\sqrt{x}+4}{\sqrt{x}-2}\)
