ĐK$x> 0$
a) rútx gọn đk $M=\frac{\sqrt{x}+1}{\sqrt{x}-1}$
b)Để $M=\sqrt{x}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x-\sqrt{x}}{\sqrt{x}-1}\Rightarrow \sqrt{x}+1=x-\sqrt{x}$$\Leftrightarrow x-2\sqrt{x}-1=0$
ra z tự lm nốt nhé !!!
c)$M=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}$
Theo bài ra : $x\in N\Rightarrow \sqrt{x}-1\in Ư(2)$
Mà $x>0\Rightarrow \sqrt{x}-1>-1$
Thay
+) $\sqrt{x}-1=1\Rightarrow x=4(thoả mãn);M=3(thoả mãn )$
+)$\sqrt{x}-1=2\Rightarrow x=9(thoả mãn );M=2(thoả mãn )$
vậy ........
~~~~~$Amen$~~~~~