1) \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}-1}{2}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
2) \(M\le\dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{2}\le0\Rightarrow\dfrac{\sqrt{x}-1}{2\sqrt{x}+2}\le0\)
mà \(2\sqrt{x}+2>0\Rightarrow\sqrt{x}-1\le0\Rightarrow\sqrt{x}\le1\Rightarrow x\le1\Rightarrow0\le x\le1\)
3) \(\dfrac{1}{M}\in Z\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}\in Z\Rightarrow1+\dfrac{1}{\sqrt{x}}\in Z\)
mà \(x\in Z\Rightarrow1⋮\sqrt{x}\Rightarrow\sqrt{x}=1\left(\sqrt{x}\ge0\right)\Rightarrow x=1\)
mà \(x\ne1\Rightarrow\) không có x nguyên để \(\dfrac{1}{M}\in Z\)
