a) \(ĐK:x\ge0,x\ne1\)

 \(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+4+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2x+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)

b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Kết hợp với đk:

\(\Rightarrow0\le x< 1\)

\(x+y+z+8=2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\left(1\right)\)

Áp dụng Bđt Bunhiacopxki :

\(\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le\left(2^2+4^2+6^2\right)\left(x-1+y-2+z-3\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z-6\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z+8\right)-784\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=\dfrac{x+y+z-6}{14}\left(2\right)\)

Đặt \(t=x+y+z+8\)

\(\left(1\right)\Leftrightarrow t^2=56t-784\)

\(\Leftrightarrow t^2-56t+784=0\)

\(\Leftrightarrow\left(t-28\right)^2=0\)

\(\Leftrightarrow t=28\)

\(\Leftrightarrow x+y+z+8=28\)

\(\Leftrightarrow x+y+z-6=14\)

\(\left(2\right)\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.4=4\\z-2=1.8=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=10\end{matrix}\right.\) thỏa mãn đề bài

<=>x-1-2\(\sqrt{x-1}+1+\sqrt{x-2}\)=0

<=>(\(\sqrt{x-1}+1\))²+\(\sqrt{x-2}\)=0

ta có \(\left(\sqrt{x-1}+1\right)^2\ge0\) với mọi x

và\(\sqrt{x-2}\ge0\) với mọi x

=>\(\left(\sqrt{x-1}+1\right)^2+\sqrt{x-2}\ge0\)

để dấu = xảy ra thì

\(\left(\sqrt{x-1}+1\right)^2=0\) =>\(\sqrt{x-1}=-1\)(loại)

và \(\sqrt{x-2}=0\)=> x=2

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\\ A\in Z\Rightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\Rightarrow\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)\\ \Leftrightarrow\left(\sqrt{x}-1\right)\in\left\{2;1;-1;-2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;2;0;-1\right\}\\ \Leftrightarrow x\in\left\{9;4;0\right\}\)

Vậy \(x\in\left\{9;4;0\right\}\)

\(P=-\left(\left(4-x\right)-\sqrt{4-x}+\dfrac{1}{4}\right)+\dfrac{17}{4}=-\left(\sqrt{4-x}-\dfrac{1}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\)

Dấu "=" xảy ra khi \(\sqrt{4-x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{15}{4}\)