Bài 1:
1/
\(A=\left[\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{x}{\sqrt{x}(\sqrt{x}-1)}\right]:\left[\frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}\right]\)
\(=\frac{x+1}{\sqrt{x}(\sqrt{x}-1)}:\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{x+1}{\sqrt{x}(\sqrt{x}-1)}.(\sqrt{x}-1)=\frac{x+1}{\sqrt{x}}\)
2/
\(A-2=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{(\sqrt{x}-1)^2}{\sqrt{x}}>0\) với mọi $x>0; x\neq 1$
Do đó ta có đpcm.
Bài 2:
1/
\(P=\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3x+3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3}{\sqrt{x}+3}\)
2/
\(P=\frac{1}{3}\Leftrightarrow \frac{3}{\sqrt{x}+3}=\frac{1}{3}\)
$\Leftrightarrow \sqrt{x}+3=9$
$\Leftrightarrow \sqrt{x}=6$
$\Leftrightarrow x=36$ (thỏa đkxđ)
Vậy.........
3/
Vì $\sqrt{x}\geq 0$ với mọi $x\in $ ĐKXĐ
$\Rightarrow \sqrt{x}+3\geq 3$
$\Rightarrow P=\frac{3}{\sqrt{x}+3}\leq 1$
Vậy $P_{\max}=1$ khi $x=0$
