\(M=\left(\dfrac{2}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-4}\right):\dfrac{\sqrt{x}}{x-4\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-8+x}{\sqrt{x}\left(\sqrt{x}-4\right)}\cdot\dfrac{x-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}-8}{\sqrt{x}}\cdot\dfrac{x-4\sqrt{x}}{x-4\sqrt{x}}=\dfrac{x+2\sqrt{x}-8}{\sqrt{x}}\)
Để 0<=M<=2 thì \(\left\{{}\begin{matrix}M>=0\\M< =2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2\sqrt{x}-8>=0\\\dfrac{x+2\sqrt{x}-8-2\sqrt{x}}{\sqrt{x}}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(\sqrt{x}+4\right)\left(\sqrt{x}-2\right)>=0\\x-8< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-2>=0\\x< =8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>=2\\x< =8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=4\\x< =8\end{matrix}\right.\)
=>4<=x<=8
mà x là số nguyên tố
nên \(x\in\left\{5;7\right\}\)
=>Có 2 giá trị x là số nguyên tố thỏa mãn
