a)
\(A=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) \(A=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)(T/m)
c) \(\dfrac{1}{A}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}}-\sqrt{x}>0\)
\(\Leftrightarrow\dfrac{-\sqrt{x}+1}{\sqrt{x}}>0\)
\(\Leftrightarrow-\sqrt{x}+1>0\left(\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
mà \(x\ge0\) mà x dưới mẫu \(\Rightarrow0< x< 1\)
