a/ ĐK: $x>0,x\ne 1$
Đặt $A=\left(\dfrac{1}{\sqrt x-1}-\dfrac{1}{\sqrt x}\right):\left(\dfrac{\sqrt x+1}{\sqrt x+3}+\dfrac{\sqrt x+3}{\sqrt x-1}\right)$
$=\left(\dfrac{\sqrt x}{\sqrt x(\sqrt x-1)}-\dfrac{\sqrt x-1}{\sqrt x(\sqrt x-1)}\right):\left(\dfrac{(\sqrt x+1)(\sqrt x-1)}{(\sqrt x+3)(\sqrt x-1)}+\dfrac{(\sqrt x+3)^2}{(\sqrt x+3)(\sqrt x-1)}\right)\\=\dfrac{\sqrt x-(\sqrt x-1)}{\sqrt x(\sqrt x-1)}:\dfrac{(\sqrt x-1)(\sqrt x+1)+(\sqrt x+3)^2}{(\sqrt x+3)(\sqrt x-1)}\\=\dfrac{\sqrt x-\sqrt x+1}{\sqrt x(\sqrt x-1)}.\dfrac{(\sqrt x+3)(\sqrt x-1)}{x-1+x+6\sqrt x+9}\\=\dfrac{1}{\sqrt x}.\dfrac{\sqrt x+3}{2x+6\sqrt x+8}\\=\dfrac{\sqrt x+3}{\sqrt x(2x+6\sqrt x+8)}$
Vậy $A=\dfrac{\sqrt x+3}{\sqrt x(2x+6\sqrt x+8)}$ với $x>0,x\ne 1$
b/ $(*)2x+6\sqrt x+8\\=2x+6\sqrt x+\dfrac{9}{2}+\dfrac{7}{2}\\=\left(2x+6\sqrt x+\dfrac{9}{2}\right)+\dfrac{7}{2}\\=2\left(x+3\sqrt x+\dfrac{9}{4}\right)+\dfrac 7 2\\=2\left(\sqrt x+\dfrac{3}{2}\right)^2+\dfrac 7 2$
Ta có: $\sqrt x> 0$ (vì $x>0$)
$→\sqrt x+\dfrac{3}{2}>0\\→\left(\sqrt x+\dfrac{3}{2}\right)^2>0\\→2\left(\sqrt x+\dfrac{3}{2}\right)^2>0\\→2\left(\sqrt x+\dfrac 3 2\right)^2+\dfrac 7 2>0\\→2x+6\sqrt x+8>0\\→\sqrt x(2x+6\sqrt x+8)>0(1)$
$(*)\sqrt x>0$ (vì $x>0$)
$→\sqrt x+3>0(2)$
(1)(2) $→\dfrac{\sqrt x+3}{\sqrt x(2x+6\sqrt x+8)}>0\\↔A>0$
mà $A<0$
$→x∈\varnothing$
Vậy không có giá trị $x$ thỏa mãn hay $x∈\varnothing$
