a, P= \(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)(đk: \(x>0,x\ne1\))
= \(\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(2+\sqrt{x}\right)}=\frac{x}{\sqrt{x}-1}\)
b,Để P>2 <=> \(\frac{x}{\sqrt{x}-1}-2>0\) <=> \(\frac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
<=>P= \(\frac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
Có \(\left(\sqrt{x}-1\right)^2+1\ge1>0\) vs mọi \(x\ge0\)
=> Để P>2 <=> \(\sqrt{x}-1>0\) <=> \(\sqrt{x}>1\) <=> \(x>1\)
Vậy để P>2 <=> x>1
