a, \(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\) (đk: \(x\ge0,x\ne1\))
= \(\left[\frac{\sqrt{x^3}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\frac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
=\(\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)^2}\)
= \(\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right).\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
=\(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
=\(\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}.2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b,Có P<0
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\) <=> \(\sqrt{x}-1< 0\) (do \(\sqrt{x}+1>0\) với mọi x t/m đk)
<=> \(\sqrt{x}< 1\) <=> x<1 .Kết hợp với đk của x => 0\(\le x< 1\)
VậyP<0 <=> \(0\le x< 1\)
c, Có P=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để \(P\in Z\) <=> \(\frac{2}{\sqrt{x}-1}\in Z\)
Với \(x\ge0,x\ne1\) =>\(\left[{}\begin{matrix}\sqrt{x}\in N\\\sqrt{x}\notin N\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}\sqrt{x}-1\in Z\\\sqrt{x}-1\notin Z\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}\frac{2}{\sqrt{x}-1}\in Z\left(tm\right)\\\frac{2}{\sqrt{x}-1}\notin Z\left(ktm\right)\end{matrix}\right.\)
=> \(\sqrt{x}-1\inƯ\left(2\right)=\left\{1,-1,2,-2\right\}\)
<=> \(\sqrt{x}\in\left\{2;0;3-2\right\}\)
mà \(\sqrt{x}\ge0\)=> \(\sqrt{x}\in\left\{2;0;3\right\}\)
<=> \(x\in\left\{4;0;9\right\}\)
Vậy để P nguyên <=> \(x\in\left\{4;0;9\right\}\)
