ĐKXĐ: \(x\ge0,x\ne1\)
a/ \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
= \(\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
= \(\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/
Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để A nhận giá trị nguyên \(\Leftrightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) ( vì \(1\in Z\) )
\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(2\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=2\\\sqrt{x}-1=-2\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\\x=0\end{matrix}\right.\) (tm)
Vậy để A nhận giá trị nguyên thì \(x=\left\{9;4;0\right\}\)
