`#tnvt`
`a,`
`P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}(x\ne1,x>=0)`
`=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`=\frac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`=\frac{\sqrt{x}-1}{\sqrt{x}+1}`
$\\$
`b,`
`P=-1`
`=>\frac{\sqrt{x}-1}{\sqrt{x}+1}=-1(x>=0,x\ne1)`
`<=>\sqrt{x}-1=-(\sqrt{x}+1)`
`<=>\sqrt{x}-1=-\sqrt{x}-1`
`<=>2\sqrt{x}=0`
`<=>\sqrt{x}=0`
`<=>x=0(tm)`
Vậy `x=0` thì `P=-1`
$\\$
`c,`
`P<2`
`=>\frac{\sqrt{x}-1}{\sqrt{x}+1}<2(x\ne1,x>=0)`
`<=>\frac{\sqrt{x}-1}{\sqrt{x}+1}-2<0`
`<=>\frac{\sqrt{x}-1-2(\sqrt{x}+1)}{\sqrt{x}+1}<0`
`<=>\frac{\sqrt{x}-1-2\sqrt{x}-2}{\sqrt{x}+1}<0`
`<=>\frac{-\sqrt{x}-3}{\sqrt{x}+1}<0`
`AAx>=0,` ta có: `\sqrt{x}+1>=1>0`
`=>-\sqrt{x}-3<0`
`<=>-\sqrt{x}<3`
`<=>\sqrt{x}> -3(\text{Luôn đúng})`
`->P<2AAx`
$\\$
`d,`
`P=\frac{\sqrt{x}-1}{\sqrt{x}+1}(x\ne1,x>=0)`
Ta xét hiệu.
`P-1`
`=\frac{\sqrt{x}-1}{\sqrt{x}+1}-1`
`=\frac{\sqrt{x}-1-(\sqrt{x}+1)}{\sqrt{x}+1}`
`=\frac{\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1]`
`=\frac{-2}{\sqrt{x}+1}`
`AAx>=0,` ta có: `\sqrt{x}+1>=1>0`
Mà `-2<0`
`=>\frac{-2}{\sqrt{x}+1}<0`
`=>P-1<0`
`=>P<1`
Vậy `P<1AAx>=0,x\ne1`
$\\$
`e,`
`P=\frac{\sqrt{x}-1}{\sqrt{x}+1}(x\ne1,x>=0)`
`=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}`
`=1-\frac{2}{\sqrt{x}+1]`
`AAx>=0` ta có: `\sqrt{x}+1>=1>0`
`=>\frac{2}{\sqrt{x}+1}<=2/1=2`
`=>1-\frac{2}[\sqrt{x}+1}>=1-2=-1`
Dấu `=` xảy ra khi `\sqrt{x}+1=1<=>x=0`
Vậy `GTNNNN_P=-1` khi `x=0`
