\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{2\sqrt{x}+1}{x+\sqrt{x}-2}\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}+1}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
b) \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{1}{\sqrt{x}+1}\le1\Rightarrow1+\dfrac{1}{\sqrt{x}+1}\le2\)
\(\Rightarrow A_{max}=2\) khi \(x=0\)
