a,ĐKXĐ: x\(\ge\)0 ; x\(\ne\)1
P=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b,để P=-1 thì :
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)=-1
<=>\(\sqrt{x}-1=-\sqrt{x}-1\)
<=>2\(\sqrt{x}\)=0
<=>x=0 (TMĐKXĐ)
c, Ta có :
P=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)=\(\text{}\text{}\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\)=1-\(\dfrac{2}{\sqrt{x}+1}\)
Để P nguyên thì :
1-\(\dfrac{2}{\sqrt{x}+1}\) nguyên
<=>\(\dfrac{2}{\sqrt{x}+1}\) nguyên
<=>2 \(⋮\) (\(\sqrt{x}+1\))
<=>\(\sqrt{x}+1\) \(\in\) Ư(2)={1;-1;2;-2}
TH1:\(\sqrt{x}+1\)=1 <=>x=0(TM)
TH2:\(\sqrt{x}+1\)=-1 <=>\(\sqrt{x}\)=-2 (ko TM)
TH3:\(\sqrt{x}+1\)=2 <=>x=1 (ko TM)
TH4:\(\sqrt{x}+1\)=-2<=>\(\sqrt{x}\)=-3 (ko TM)
Vậy để P nguyên thì x=1
