ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
a) \(A=\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right):\left(\dfrac{x+\sqrt{x}-4}{x-2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\right)=\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}:\left[\dfrac{x+\sqrt{x}-4}{x+\sqrt{x}-3\sqrt{x}-3}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right]=\dfrac{1}{\sqrt{x}-2}:\left[\dfrac{x+\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}-\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right]=\dfrac{1}{\sqrt{x}-2}:\left[\dfrac{x+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right]=\dfrac{1}{\sqrt{x}-2}:\dfrac{x+\sqrt{x}-4-x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}:\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}:\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)b) Ta có A=2\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=2\Leftrightarrow\sqrt{x}+1=2\left(\sqrt{x}-2\right)\Leftrightarrow\sqrt{x}+1=2\sqrt{x}-4\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)Vậy x=25 thì A=2
c) Ta có \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)
Vậy để A nhận giá trị nguyên thì \(\sqrt{x}-2\inƯ\left(3\right)\in\left\{\pm1;\pm3\right\}\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}-2=1\\\sqrt{x}-2=-1\\\sqrt{x}-2=3\\\sqrt{x}-2=-3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=5\\\sqrt{x}=-1\left(l\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=9\left(ktm\right)\\x=1\left(tm\right)\\x=25\left(tm\right)\end{matrix}\right.\)
Vậy x=1 hoặc x=25 thì A nhận giá trị nguyên
