a) điều kiện \(x\ge0;x\ne4;x\ne9\)
\(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{2\sqrt{x}-9-\left(x-9\right)+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) ta có : \(Q=0\) \(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=9\left(loại\right)\end{matrix}\right.\) vậy không có giá trị nào của x để Q = 0
c) ta có : \(x=\sqrt{7+\sqrt{24}}\Leftrightarrow x=\sqrt{\left(\sqrt{6}+1\right)^2}\Leftrightarrow x=\sqrt{6}+1\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}+1\right)}\)
thay vào Q ta có \(Q=\dfrac{\sqrt{\left(\sqrt{6}+1\right)}+1}{\sqrt{\left(\sqrt{6}+1\right)}-3}\)
d) ta có : \(Q>0\) \(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\)
mà ta có : \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1>0\)
\(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\Leftrightarrow\sqrt{x}-3>0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
vậy \(x\ge9\) thì \(Q>0\)
