a: ĐKXĐ: x>=0; x<>1

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}+1\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

b: Thay \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\) vào P, ta được:

\(P=\frac{\sqrt{\left(\sqrt2+1\right)^2}+2}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+2}{\sqrt2+1-1}=\frac{3+\sqrt2}{\sqrt2}=\frac{3\sqrt2+2}{2}\)

c: P<0

=>\(\frac{\sqrt{x}+2}{\sqrt{x}-1}<0\)

=>\(\sqrt{x}-1<0\)

=>\(\sqrt{x}<1\)

=>0<=x<1

\(y=\sqrt{x-1}+\sqrt{9-x}\)(đk: \(9\ge x\ge1\))
=> \(y\ge\sqrt{x-1+9-x}=\sqrt{8}\)
Dấu "=" xảy ra khi x =1 hoặc x= 9 

Vậy y min  = \(\sqrt{8}\)khi x =1 hoặc x = 9

\(P=\dfrac{x-1+2}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{2}{\sqrt{x}-1}\)

\(=\sqrt{x}-1+\dfrac{2}{\sqrt{x}-1}+2>=2\sqrt{2}+2\)

Dấu = xảy ra khi (căn x-1)^2=2

=>(căn x-1)=căn 2

=>căn x=căn 2+1

=>x=3+2 căn 2

\(A=x+\frac{2}{\sqrt{x}}\)

\(=x+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\)

\(\ge3\sqrt[3]{x\cdot\frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x}}}=3\)

Dấu "=" xảy ra tại x=1

Vậy \(GTLN_A=3\Leftrightarrow x=1\)