\(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}-1}{2}\left(x\ge0,x\ne1\right)\)
\(M=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2}\)
\(M=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right).2}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
2, \(M\le\dfrac{1}{2}=>\dfrac{\sqrt{x}}{\sqrt{x}+1}\le\dfrac{1}{2}< =>\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{2}\le0\)
\(< =>\dfrac{2\sqrt{x}-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}\le0\)
\(< =>\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}\le0< =>\sqrt{x}-1\le0\)
\(< =>\sqrt{x}\le1< =>x\le1\)
kết hợp với điều kiện xác định=>\(0\le x\le1\) thì M\(\le\dfrac{1}{2}\)
3.\(\dfrac{1}{M}=\dfrac{1}{\dfrac{\sqrt{x}}{\sqrt{x}+1}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}=1+\dfrac{1}{\sqrt{x}}\)
để \(\dfrac{1}{M}\) nguyên khi \(1+\dfrac{1}{\sqrt{x}}\) nguyên
=>\(\sqrt{x}\inƯ\left(1\right)=\left\{1,-1\right\}\)
với \(\sqrt{x}=1=>x=1\left(Loai\right)\)
với \(\sqrt{x}=-1\)(vô lí)
=>x\(\in\phi\)
