$a)$ĐKXĐ: $x\geq0; x\neq4$
$M=(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}):\frac{2}{\sqrt{x}-2}$
$M=(\frac{\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}+\frac{1}{\sqrt{x}-2}).\frac{\sqrt{x}-2}{2}$
$M=\frac{\sqrt{x}+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{2}$
$M=\frac{2\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{2}$
$M=\frac{2(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{2}$
$M=\frac{\sqrt{x}+1}{\sqrt{x}+2}$
Vậy với $x\geq0; x\neq4$ thì $M=\frac{\sqrt{x}+1}{\sqrt{x}+2}$
$b)$Để $M=\frac{4}{5}$
`<=>`$\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{4}{5}$
`<=>`$5(\sqrt{x}+1)=4(\sqrt{x}+2)$
`<=>`$5\sqrt{x}+5=4\sqrt{x}+8$
`<=>`$\sqrt{x}=3$
`<=>`$x=9 (t/m)$
Vậy $x=9$ thì $M=\frac{4}{5}$
$c)M=\frac{\sqrt{x}+1}{\sqrt{x}+2}\geq\frac{1}{2}>0$ với $∀x\leq0; x\neq4$
Bên cạnh đó:
$M=\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\leq1-\frac{1}{2}<1$ với $∀x\leq0; x\neq4$
`=>`$M-1<0$
`=>`$M(M-1)<0$
`<=>`$M^{2}-M<0$
`<=>`$M^{2}<M$
Vậy $M>M^{2}$
