ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Câu a : \(M=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x\sqrt{x}-1}\right).\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{x+\sqrt{x}+1}\)
Câu b : Để \(M\in Z\) \(\Leftrightarrow\dfrac{3}{x+\sqrt{x}+1}\in Z\)
Thì : \(x+\sqrt{x}+1\inƯ\left(3\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+\sqrt{x}+1=-3\\x+\sqrt{x}+1=-1\\x+\sqrt{x}+1=1\\x+\sqrt{x}+1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x\in\varnothing\\x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy không có giá trị của x để M nguyên !
