$M\,=A:B\\\quad =\dfrac{x-\sqrt x+2}{\sqrt x+3}:\dfrac{\sqrt x-1}{\sqrt x+3}\\\quad =\dfrac{x-\sqrt x+2}{\sqrt x+3}.\dfrac{\sqrt x+3}{\sqrt x-1}\\\quad =\dfrac{x-\sqrt x+2}{\sqrt x-1}\\\quad =\dfrac{(x-\sqrt x)+2}{\sqrt x-1}\\\quad =\sqrt x+\dfrac{2}{\sqrt x-1}\\\quad =(\sqrt x-1)+\dfrac{2}{\sqrt x-1}+1$
Vì $x>1$
$→\sqrt x>1\\↔\sqrt x-1>0\\→\dfrac{2}{\sqrt x-1}>0$
Áp dụng bất đẳng thức Cô-si với 2 số dương $\sqrt x-1,\,\dfrac{2}{\sqrt x-1}$
$(\sqrt x-1)+\dfrac{2}{\sqrt x-1}\ge 2\sqrt{(\sqrt x-1).\dfrac{2}{\sqrt x-1}}\\↔(\sqrt x-1)+\dfrac{2}{\sqrt x-1}\ge 2\sqrt 2\\→M\ge 2\sqrt 2+1$
$→$ Dấu "=" xảy ra khi $\sqrt x-1=\dfrac{2}{\sqrt x-1}$
$↔\dfrac{(\sqrt x-1)^2}{\sqrt x-1}=\dfrac{2}{\sqrt x-1}\\→(\sqrt x-1)^2=2\\↔\sqrt x-1=\sqrt 2(vì\,\,\sqrt x-1>0)\\↔\sqrt x=1+\sqrt 2\\↔x=3+2\sqrt 2(TM)$
Vậy $M$ đạt GTNN là $2\sqrt 2+1$ khi $x=3+2\sqrt 2$
