ĐK: $x\ge 0,x\ne 1,x\ne \dfrac{1}{4}$

a, $A=\bigg(\dfrac{\sqrt x+1}{\sqrt x-1}+\dfrac{\sqrt x-1}{\sqrt x+1}-\dfrac{3\sqrt x+1}{x-1}\bigg).\dfrac{3}{2\sqrt x-1}\\=\bigg(\dfrac{\sqrt x+1}{\sqrt x-1}+\dfrac{\sqrt x-1}{\sqrt x+1}-\dfrac{3\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}\bigg).\dfrac{3}{2\sqrt x-1}\\=\bigg(\dfrac{(\sqrt x+1)^2}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{(\sqrt x-1)^2}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{3\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}\bigg).\dfrac{3}{2\sqrt x-1}\\=\dfrac{x+2\sqrt x+1+x-2\sqrt x+1-3\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{3}{2\sqrt x-1}\\=\dfrac{2x-3\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{3}{2\sqrt x-1}\\=\dfrac{(2\sqrt x-1)(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{3}{2\sqrt x-1}\\=\dfrac{3}{\sqrt x+1}$

Vậy ta có ĐPCM

2, $A\le 1$ hay $\dfrac{3}{\sqrt x+1}\le 1$

$↔\dfrac{3}{\sqrt x+1}-1\le 0\\↔\dfrac{3-\sqrt x-1}{\sqrt x+1}\le 0\\↔\dfrac{2-\sqrt x}{\sqrt x+1}\le 0\\↔2-\sqrt x\le 0(do\,\,\sqrt x+1>0\forall x\ge 0,x\ne 1,x\ne \dfrac{1}{4})\\↔\sqrt x\ge 2\\↔x\ge 4$

(Câu này hình như đề sai, $A\ge 1$ mới đúng)

Sử $A\ge 1$

$A\ge 1$ hay $\dfrac{3}{\sqrt x+1}\ge 1$

$↔\dfrac{3}{\sqrt x+1}-1\ge 0\\↔\dfrac{2-\sqrt x}{\sqrt x+1}\ge 0\\↔2-\sqrt x\ge 0(do\,\,\sqrt x+1>0\forall x\ge 0,x\ne 1,x\ne \dfrac{1}{4})\\↔\sqrt x\le 2\\↔x\le 4$

Kết hợp điều kiện $x\ge 0,x\ne 1,x\ne \dfrac{1}{4}$

$→\begin{cases}0\le x\le 4\\x\ne 1;x\ne \dfrac{1}{4}\end{cases}$

Mà $x$ nguyên lớn nhất

$→x=4(TM)$

Vậy $x=4$