a) \(E=\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\dfrac{x}{\sqrt{x}-1}\)
b) \(E>1\) khi:
\(\dfrac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
Mà: \(x-\sqrt{x}+1>0\forall x\)
\(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow x>1\)
c) Ta có:
\(E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1-2\sqrt{x}+2\sqrt{x}}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2+2\sqrt{x}-2+1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\dfrac{2\left(\sqrt{x}-1\right)+1}{\sqrt{x}-1}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
Theo BĐT cô-si ta có:
\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}=2\)
\(\Rightarrow E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2+2=4\)
Vậy: \(E_{min}=4\)
d) \(E=\dfrac{9}{2}\) khi:
\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)
\(\Leftrightarrow2x=9\sqrt{x}-9\)
\(\Leftrightarrow2x-9\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{9}{4}\end{matrix}\right.\)
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