Bài 1:
PT $\Leftrightarrow (\sqrt{2}+1)x=2+\sqrt{2}$
$\Leftrightarrow x=\frac{2+\sqrt{2}}{\sqrt{2}+1}=\frac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1}=\sqrt{2}$
Vậy pt có nghiệm $x=\sqrt{2}$
Bài 2:
\(P=\left[\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}-\frac{1}{\sqrt{x}(\sqrt{x}+1)}\right]:\frac{\sqrt{x}-1}{(\sqrt{x}+1)^2}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}+1)}.\frac{(\sqrt{x}+1)^2}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)