A= \(\frac{x-\sqrt{x}}{x-1}\left(đk:x\ge0,x\ne1\right)\)
= \(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)(vì \(x\ge0\))
=\(\frac{\sqrt{x}}{\sqrt{x}+1}\)
B=\(\frac{x-4}{x-2\sqrt{x}}\) (đk: \(x>0,x\ne4\))
= \(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=1+\frac{2}{\sqrt{x}}\)
b, Có A=\(\frac{\sqrt{x}}{\sqrt{x}+1}< 1\)(vì \(x\ge0\))
B=\(1+\frac{2}{\sqrt{x}}>1\) (vì \(x\ge0\))
=> A<B
c,Có A.B=\(\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)
Để A.B \(\in Z\) => \(\frac{1}{\sqrt{x}+1}\in Z\)
Với x\(\ge0\) có \(\left[{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\notin Z\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\sqrt{x}+1\in Z\\\sqrt{x}+1\notin Z\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\frac{1}{\sqrt{x}+1}\in Z\left(tm\right)\\\frac{1}{\sqrt{x}+1}\notin Z\left(ktm\right)\end{matrix}\right.\)
=> \(\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
<=> \(\sqrt{x}\in\left\{0,-2\right\}\)
mà \(\sqrt{x}\ge0\)=> \(\sqrt{x}=0\) <=> x=0(t/m)
Vậy để A.B \(\in Z\) <=> x=0
d,Có A.B< \(\frac{1}{2}\)
<=>\(1+\frac{1}{\sqrt{x}+1}< \frac{1}{2}\) <=> \(\frac{1}{\sqrt{x}+1}+\frac{1}{2}< 0\)(vô lý)
Vậy không có x t/m A.B<\(\frac{1}{2}\)
