Giải thích các bước giải:
1.ĐKXĐ: $x\ge 0, x\ne 1$
Ta có:
$P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}$
$\to P=\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{6\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to P=\dfrac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)-(6\sqrt{x}-4)}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to P=\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to P=\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
2.Để $P<0$
$\to \dfrac{\sqrt{x}-1}{\sqrt{x}+1}<0$
Mà $\sqrt{x}+1>0$
$\to \sqrt{x}-1<0$
$\to \sqrt{x}<1$
$\to 0\le x<1$
3.Ta có:
$P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
$\to P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}$
$\to P=1-\dfrac{2}{\sqrt{x}+1}$
$\to P\ge 1-\dfrac{2}{0+1}$
$\to P\ge -1$
$\to GTNN_p=-1$ khi đó $x=0$
4.Để $P=-\sqrt{x}$
$\to \dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-\sqrt{x}$
$\to \sqrt{x}-1=-\sqrt{x}(\sqrt{x}+1)$
$\to \sqrt{x}-1=x-\sqrt{x}$
$\to x-2\sqrt{x}+1=0$
$\to (\sqrt{x}-1)^2=0$
$\to \sqrt{x}-1=0$
$\to \sqrt{x}=1$
$\to x=1$
Mà $x\ne 1\to $Không tồn tại $x$ thỏa mãn đề
5.Ta có:
$P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
$\to P-1=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-1$
$\to P-1=\dfrac{\sqrt{x}-1-(\sqrt{x}+1)}{\sqrt{x}+1}$
$\to P-1=\dfrac{-2}{\sqrt{x}+1}<0$ vì $\sqrt{x}+1>0$
$\to P<1$
